A third revision of the article is now on arXiv. --2004-Nov-27
Some solutions can be found at http://www.ark.in-berlin.de/conj-sol.pdf
Do you have a comment or news on conjectures in the article math.CO/0409509?
Please mail to ralf@ark.in-berlin.de
Last changes: 2005-Oct-25
1 (A002421) [x^n](1-4x)sqrt(1-4x) = C(2n,n)/(1-2n) - 4*C(2n-2,n-1)/(3-2n). --RS, 2004-Sep-27
2 (A071721) Solution through [x^n](-1+x)sqrt(1-4x) = -C(2n,n)/(1-2n) + C(2n,n)/(3-2n). --RS, 2004-Sep-27
3 (A004130) Proved by Mitch Harris, 2004-Oct-11
4 (A001800) Eqn 28 of http://mathworld.wolfram.com/LegendrePolynomial.html. --researched by Mitch Harris, 2004-Oct-12
5 (A077071) Proved by Nikolaus Meyberg, 2004-Oct-15
6 (A001896) Consequence of the stronger statement that B(2n,1/2) = 2^{2n}B(2n,1/4), proved via g.f., together with the fact that the numerator of B(2n, 1/2) is odd. --Ira Gessel, 2004-Sep-29
7 (A012670) Proved by Robin Chapman, pers. comm., 2004-Sep-30, and Seunghyun Seo, 2004-Oct-02
8 (A035174) Follows from the formula tau(p^{n+2}) = tau(p)tau(p^{n+1}) - p^11 tau(p^n) for prime p, which comes from the theory of Hecke operators on modular forms. The p = 2 case gives a recurrence for tau(2^n) leading immediately to (8). --Robin Chapman, 2004-Sep-29
9 (A094088) For any odd prime p, the coefficients of 1/(2-cosh(x)) as e.g.f. are periodic with period dividing p-1. --Ira Gessel, 2004-Sep-29
10 (A005046) Follows also from a very general divisibility property of such function expansions, as shown by Ira Gessel, 2004-Oct-02
11 (A085435) Resultant(p,q) = (Leading Coefficient of q)^(Degree of p) * Product(p(i):i roots of q) = 4^n*(1/2^n-1)*((-1/2)^n-1) = r.h.s. --Dr. Luke Pebody, 2004-Oct-12
12 (A004782) Proved by Dr. Luke Pebody. --2004-Oct-12
13 (A026937) Proved by Ira Gessel. --2004-Oct-03
14 (A027305) This is Sum[k=0..n+1, (k+1) C(2n+1,k)] - Sum[k=1..n+1, (k+1) C(2n+1,k-1)] = - Sum[k=0..n, C(2n+1,k)] + (n+2) C(2n+1,n+1) = (-1/2) (1+1)^(2n+1) + (n+2) C(2n+1,n+1) = -4^n + (n+2)/2 C(2n+2,n+1). ( recurrence: a(n) = 4 a(n-1) + binomial(2n,n-1) ) --Vladeta Jovovic, 2004-Oct-09
15 (A001317) Follows from a(n)=\product_{k \in K} (1+2^(2^k)), where K is the set of integers such that n=\sum_{k \in K} 2^k. --Emmanuel Ferrand, 2004-Sep-28
16 (A099173) Multiply the left side by x^n and sum on n (using the binomial theorem), then sum on k (geometric series). --Ira Gessel, 2004-Sep-29
17 (A080170)
18 (A031698) Disproved (first counterexample is 48862) by Luke Pebody who gave even a general form of the exceptions, 2004-Oct-15.
19 (A065304) Proved by Matthew Akeran using that n is the sum of two coprime square numbers. --2005-Jan-26
20 (A077626) The largest term in the p.p. of the c.f. of sqrt(n) is 2*floor(sqrt(n)), --Luke Pebody, 2004-Oct-15
21 (A041042) Labelling the numerators of the continued fraction of sqrt(27) a_1, b_1, a_2, b_2, a_3, b_3, ..., we get the mutual recurrences a_(n+1) = 10*b_n + a_n, b_(n+1) = 5*a_(n+1) + b_n = 51*b_n + 5*a_n, giving a transition matrix from (a_n b_n) to (a_(n+1) b_(n+1)) of ( 1 10 ) ( 5 51 ). --Excerpt from proof by Timothy Y. Chow, 2004-Sep-29
22 (A056739) General case n=p^iq^i => n|Sum[k=1..2p, k^n], where p, q=2p+1 prime, proved by Christian Meyer, 2004-Oct-01, see http://enriques.mathematik.uni-mainz.de/cm/article_primes.pdf
23 (A062960) Exponent of largest prime dividing n! in prime factorization of n! is 1, i.e., n! = p_1^e_1*p_2^e_2*...*p_(s-1)^e_(s-1)*p_s, p_1=5.
50 (A003668) Proved by Matthew Akeran, see http://www.ark.in-berlin.de/A003668.pdf , 2004-Nov-15
51 (A000404, A084889) Proved by Jon Namnath and Lawrence Sze, 2004-Oct-22
52 (A026471) Proved by Matthew Akeran, 2004-Nov-04
53 (A026472) Proved by Lawrence Sze, 2004-Nov-08
54 (A026473) Proved by Lawrence Sze, 2004-Nov-08
55 (A026475) Proved by Lawrence Sze, 2004-Nov-08
56 (A026476) Proved by Lawrence Sze, 2004-Nov-08
57 (.......) Proved by Matthew Akeran, 2004-Nov-05
58 (A034702) Proved by Christian Meyer, 2004-Dec-08, see http://enriques.mathematik.uni-mainz.de/cm/article_sequence.pdf
59 (A075123) Proved by Marc Schwartz, 2004-Nov-07 and Lawrence Sze, 2004-Dec-01
60 (A004793) Proved by Lawrence Sze, 2004-Nov-12
61 (A033157) Proved by Lawrence Sze, 2004-Nov-12
62 (A093678) Proved by Lawrence Sze, 2004-Nov-12
63 (A093679) Proved by Lawrence Sze, 2004-Nov-12
64 (A093680) Proved by Lawrence Sze, 2004-Nov-12
65
66
67
68 (A002820) Proved by Kent Morrison, 2004-Oct-19 see http://www.calpoly.edu/~kmorriso/Research/mnev01.pdf
69
70 (A007476) Proved by Lawrence Sze, see http://lsze.cosam.calpoly.edu/oeis70.pdf , 2004-Nov-03
71 (A022445) Proved by Matthew Akeran, 2004-Dec-24
72 (A032087) Proved by Elizabeth Wilmer, see http://www.oberlin.edu/math/faculty/wilmer/OEISconj727374.pdf , 2004-Oct-05
73 (A032091) Proved by Elizabeth Wilmer, see http://www.oberlin.edu/math/faculty/wilmer/OEISconj727374.pdf , 2004-Oct-05
74 (A032106) Proved by Elizabeth Wilmer, see http://www.oberlin.edu/math/faculty/wilmer/OEISconj727374.pdf , 2004-Oct-05
75 (A032191) Proved by Christian Meyer, 2004-Dec-28, see http://enriques.mathematik.uni-mainz.de/cm/article_necklaces.pdf
76 (A033441) Proved by Christian Meyer, 2004-Dec-29, see http://enriques.mathematik.uni-mainz.de/cm/article_turan.pdf
77 Proved by Chris Burns and Benjamin Purcell, see http://www.oberlin.edu/math/Research/Burns-Purcell.pdf
78 (A047171) Write down the number of such subsets with k elements <= (n-1)/2 as a product of two binomial coefficients, then evaluate the sum using Vandermonde's theorem. --Ira Gessel, 2004-Sep-29
79 (A053195) Follows from the fact that for odd n, level permutations of degree n are just permutations that have odd order, i.e., A053195(2*n+1) = A000246(2*n+1). --Vladeta Jovovic, 2004-Sep-29
80 (A055278) Proved by Lawrence Sze, 2004-Dec-14
81 (A069306) Proved and shown to be P(n+1) by Tomislav Doslic, 2004-Oct-22
82 (A073948) Proved by Lawrence Sze, 2004-Oct-24.
83 (A073995) Proved by Lawrence Sze, 2004-Oct-24.
84
85
86 (A091468) For definition of octopus, see sequence in OEIS or e.g. http://math.ucr.edu/home/baez/qg-spring2004/octopus.pdf
87 (A094954) Proved by Elizabeth Wilmer, see http://www.oberlin.edu/math/faculty/wilmer/OEISconj87.pdf , 2004-Oct-16
88 (A020718) Luke Pebody sent a proof for this at 2004-Oct-13, and Matthew Akeran gave an outline of proof for 88-92 at 2004-Dec-06. Although I lack the detailed analytic knowledge necessary to judge both efforts, I will count 88-92 as proved.
89 (A020746) Proved (see above).
90 (A020748) Proved (see above).
91 (A021008) Proved (see above).
92 (A021011) Proved (see above).
93 (A072222) Should read a(n+3) = 2+a(n). Proved by Lawrence Sze, 2004-Dec-12.
94 (A016131) Proved by Mitch Harris. --2004-Oct-13
95 (A074355) Proved by Mitch Harris, 2004-Dec-14
96 (A020957) Proved by Lawrence Sze, 2004-Dec-08
97 (A024551) Proved by Luke Pebody. --2004-Oct-13
98 (A082126,A082128) Both parts proved by Luke Pebody. --2004-Oct-13
99 (A082909) Proved using Kummer's Theorem by Marc Schwartz, 2004-Nov-29
100 (A001969) "Evil Numbers" recurrence proved by Luke Pebody. --2004-Oct-13
101 (A000069) Proved by Luke Pebody. --2004-Oct-13
102 (A004762) Proved by Lawrence Sze, see http://lsze.cosam.calpoly.edu/sequence102.pdf , 2004-Oct-21
103 (A005590) Proved by Matthew Akeran, 2004-Dec-12
104 (A020330) Should read '+1' at end. Proof: a(n) is simply n + n*2^[lg(n)+1]. --RS, 2004-Dec-07
105 (A004443) Should read 'n XOR m'. Shown to be true by Matthew Akeran, 2004-Dec-31
106 (A038554) Proved by Nikolaus Meyberg, 2004-Oct-20
107 (A006581) Proved by Luke Pebody. --2004-Oct-15
108 (A006582) Proved. --RS, 2004-Dec-05
109 (A006583) Proved. --RS, 2004-Oct-19
110 (A080572) Proved by Jeremy Dover, 2004-Dec-08 see http://www.ark.in-berlin.de/A0220905.pdf
111
112
113
114 (A084091) Proved by Nikolaus Meyberg. --2004-Oct-20
115 (A022905/7/8,A000123) Proved by Jeremy Dover, 2004-Dec-08 see http://www.ark.in-berlin.de/A0220905.pdf
116 (A027615,A072894,A099389) Proved by Nikolaus Meyberg, 2004-Oct-16
117