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A329073
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a(n) = (1/n)*Sum_{k=0..n-1} (40k+13)*(-1)^k*50^(n-1-k)*T_k(4,1)*T_k(1,-1)^2, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.
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4
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13, 219, 7858, 221525, 9253710, 375158958, 16882409364, 736344816813, 32964312771550, 1471835619627770, 66910145732699964, 3061043035494001682, 141458526138008430124, 6567714993530314856700, 306628434270114823521000, 14370411994543866356077725, 676259546148988495771751550
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OFFSET
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1,1
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COMMENTS
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Conjecture 1: (i) a(n) is a positive integer for each n > 0; also, a(n) is odd if and only if n is a power of two. Moreover, we have the identity Sum_{k>=0} ((40k+13)/(-50)^k)*T_k(4,1)*T_k(1,-1)^2 = 55*sqrt(15)/(9*Pi).
(ii) Let p > 5 be a prime. Then Sum_{k=0..p-1} ((40k+13)/(-50)^k)*T_k(4,1)* T_k(1,-1)^2 == (p/3)*(12 + 5*Leg(3/p) + 22*Leg(p/15)) (mod p^2), where Leg(a/p) denotes the Legendre symbol. Also, for the sum S(p) = Sum_{k=0..p-1} T_k(4,1)* T_k(1,-1)^2/(-50)^k, if Leg(-5/p) = -1 then S(p) == 0 (mod p^2); if p == 1,9 (mod 20) and p = x^2 + 5*y^2 with x and y integers then S(p) == 4x^2-2p (mod p^2); if p == 3,7 (mod 20) and 2p = x^2 + 5*y^2 with x and y integers then S(p) == 2x^2-2p (mod p^2).
Conjecture 2: (i) For any n > 0, the number b(n):=(1/n)*Sum_{k=0..n-1} (40k+27)*(-6)^(n-1-k)*T_k(4,1)*T_k(1,-1)^2 is an integer. Moreover, b(n) is odd if and only if n is a power of two.
(ii) Let p > 3 be a prime. Then Sum_{k=0..p-1} ((40k+27)/(-6)^k)*T_k(4,1)* T_k(1,-1)^2 == (p/9)*(55*Leg(-5/p) + 198*Leg(3/p)-10) (mod p^2). Also, for the sum T(p) = Sum_{k=0..p-1} T_k(4,1)*T_k(1,-1)^2/(-6)^k, if Leg(-5/p) = -1 then T(p) == 0 (mod p^2); if p == 1,9 (mod 20) and p = x^2 + 5*y^2 with x and y integers then T(p) == Leg(p/3)*(4x^2-2p) (mod p^2); if p == 3,7 (mod 20) and 2p = x^2 + 5*y^2 with x and y integers then T(p) == Leg(p/3)(2p-2x^2) (mod p^2).
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LINKS
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Zhi-Wei Sun, On sums related to central binomial and trinomial coefficients, in: M. B. Nathanson (ed.), Combinatorial and Additive Number Theory: CANT 2011 and 2012, Springer Proc. in Math. & Stat., Vol. 101, Springer, New York, 2014, pp. 257-312. Also available from arXiv:1101.0600 [math.NT], 2011-2014.
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EXAMPLE
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a(1) = 13 since (40*0+13)*(-1)^0*50^(1-1-0)*T_0(4,1)*T_0(1,-1)^2/1 = 13/1 = 13.
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MATHEMATICA
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T[b_, c_, 0]=1; T[b_, c_, 1]=b;
T[b_, c_, n_]:=T[b, c, n]=(b(2n-1)T[b, c, n-1]-(b^2-4c)(n-1)T[b, c, n-2])/n;
a[n_]:=a[n]=Sum[(40k+13)(-1)^k*50^(n-1-k)*T[4, 1, k]*T[1, -1, k]^2, {k, 0, n-1}]/n;
Table[a[n], {n, 1, 20}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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