A328786 -id:A328786 - cp's OEIS frontend

A329105 a(n) = (Sum_{k=0..n-1} (1435k+113)3240^(n-1-k)T_k(7,1)T_k(10,10)^2)/(n10^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

113, 72486, 22959360, 6667719680, 1907342690028, 546566353351560, 157644511058113920, 45818502502241488320, 13418569988503429983660, 3957929725047766692949256, 1175070020071246825359359232, 350933963579387124964687828224, 105365902497675176184788931496400, 31787539718100094004136084118514400

Offset: 1

Zhi-Wei Sun, Nov 04 2019

nonn

Conjecture 1: (i) a(n) is an integer for each n > 0. Moreover, we have Sum_{k>=0}((1435*k+113)/3240^k)*T_k(7,1)*T_k(10,10)^2 = 1452*sqrt(5)/Pi.
(ii) Let p > 3 be a prime. Then Sum_{k=0..p-1}((1435*k+113)/3240^k)*T_k(7,1)*T_k(10,10)^2 == p/9*(2420*Leg(-5/p) + 105*Leg(5/p) - 1508) (mod p^2), where Leg(a/p) denotes the Legendre symbol. For the sum S(p) = Sum_{k=0..p-1}T_k(7,1)*T_k(10,10)^2/3240^k, if Leg(-15/p) = -1 then S(p) == 0 (mod p^2); if p == 1,4 (mod 15) and p = x^2 + 15*y^2 with x and y integers then S(p) == 4*x^2-2p (mod p^2); if p == 2,8 (mod 15) and p = 3*x^2 + 5*y^2 with x and y integers then S(p) == 12*x^2-2p (mod p^2).
Conjecture 2: (i) For each n > 0, the number (Sum_{k=0..n-1}(1435*k+1322)*50^(n-1-k)*T_k(7,1)*T_k(10,10)^2)*3/(2n*10^(n-1)) is an integer.
(ii) Let p > 5 be a prime. Then Sum_{k=0..p-1}((1435*k+1322)/50^k)*T_k(7,1)*T_k(10,10)^2 == p/3*(3432*Leg(5/p) + 968*Leg(-1/p) - 434) (mod p^2). For the sum T(p) = Sum_{k=0..p-1}T_k(7,1)*T_k(10,10)^2/50^k, if Leg(-15/p) = -1 then T(p) == 0 (mod p^2); if p == 1,4 (mod 15) and p = x^2 + 15*y^2 with x and y integers then T(p) == 4*x^2-2p (mod p^2); if p == 2,8 (mod 15) and p = 3*x^2 + 5*y^2 with x and y integers then T(p) == 2p-12*x^2 (mod p^2).

			a(1) = 113 since ((1435*0+113)*3240^(1-1-0)*T_0(7,1)*T_0(10,10)^2)/(1*10^(1-1)) = 113.

Zhi-Wei Sun, Table of n, a(n) for n = 1..70
Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
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.

Cf. A104454, A328786, A329073, A329107.

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[(1435k+113)T[7,1,k]T[10,10,k]^2*3240^(n-1-k),{k,0,n-1}]/(n*10^(n-1));
Table[a[n],{n,1,14}]

A329107 a(n) = (Sum_{k=0..n-1}(840k+197)(-1)^k2430^(n-1-k)T_k(8,1)T_k(5,-5)^2)/(n5^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

Original entry on oeis.org

197, 27131, 9162090, 3337679905, 1300603606702, 526423563257310, 219304133423593380, 93259079677243221345, 40287972095635400291790, 17621949843841860346761946, 7785698346885200295051911308, 3468528609790576968835453926954, 1556035297261133424003013368953900

Offset: 1

Zhi-Wei Sun, Nov 04 2019

nonn

Conjecture: (i) a(n) is a positive integer for each n > 0. Moreover, we have Sum_{k>=0} ((840*k+197)/(-2430)^k)*T_k(8,1)*T_k(5,-5)^2 = 189*sqrt(15)/(2*Pi).
(ii) If p > 3 is a prime, then Sum_{k=0..p-1}((840*k+197)/(-2430)^k)*T_k(8,1)*T_k(5,-5)^2 == p*(52 + 5*Leg(15/p) + 140*(-15/p)) (mod p^2), where Leg(a/p) denotes the Legendre symbol.
(iii) Let p > 7 be a prime and set S(p) = Sum_{k=0..p-1}T_k(8,1)*T_k(5,-5)^2/(-2430)^k. If Leg(-105,p) = -1, then S(p) == 0 (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/7) = 1 and p = x^2 + 105*y^2 (with x and y integers), then S(p) == 4*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/7) = 1, Leg(p/3) = Leg(p/5) = -1 and 2p = x^2 + 105*y^2, then S(p) == 2*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/7) = -1 and p = 3*x^2 + 35*y^2, then S(p) == 12*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/7) = -1, Leg(p/3) = Leg(p/5) = 1, and 2p = 3*x^2 + 35*y^2, then S(p) == 6*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = 1, Leg(p/3) = Leg(p/7) = -1, and p = 5*x^2 + 21*y^2, then S(p) == 2p-20*x^2 (mod p^2). If Leg(-1/p) = Leg(p/3) = 1, Leg(p/5) = Leg(p/7) = -1, and 2p = 5*x^2 + 21*y^2, then S(p) == 2p-10*x^2 (mod p^2). If Leg(-1/p) = Leg(p/5) = -1, Leg(p/3) = Leg(p/7) = 1, and p = 7*x^2 + 15*y^2, then S(p) == 28*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = -1, Leg(p/5) = Leg(p/7) = 1, and 2p = 7*x^2 + 15*y^2, then S(p) == 14*x^2-2p (mod p^2).
One can easily check the identity in part (i) as the series converges very fast. Note also that the imaginary quadratic field Q(sqrt(-105)) has class number 8.

			a(1) = 197 since (840*0+197)*T_0(8,1)*T_0(5,-5)^2*(-1)^0*2430^(1-1-0)/(1*5^(1-1)) = 197.

Zhi-Wei Sun, Table of n, a(n) for n = 1..70
Zhi-Wei Sun, List of conjectural series for powers of Pi and other constants, arXiv:1102.5649 [math.CA], 2011-2014.
Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
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.

Cf. A328786, A329073, A329105.

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[(840k+197)T[8,1,k]T[5,-5,k]^2*(-1)^k*2430^(n-1-k),{k,0,n-1}]/(n*5^(n-1));
Table[a[n],{n,1,13}]

cp's OEIS Frontend

A329105 a(n) = (Sum_{k=0..n-1} (1435k+113)3240^(n-1-k)T_k(7,1)T_k(10,10)^2)/(n10^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

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A329107 a(n) = (Sum_{k=0..n-1}(840k+197)(-1)^k2430^(n-1-k)T_k(8,1)T_k(5,-5)^2)/(n5^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

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cp's OEIS Frontend

A329105 a(n) = (Sum_{k=0..n-1} (1435*k+113)*3240^(n-1-k)*T_k(7,1)*T_k(10,10)^2)/(n*10^(n-1)), 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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Comments

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Mathematica

A329107 a(n) = (Sum_{k=0..n-1}(840*k+197)(-1)^k*2430^(n-1-k)*T_k(8,1)*T_k(5,-5)^2)/(n*5^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+b*x+c)^k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A329105 a(n) = (Sum_{k=0..n-1} (1435k+113)3240^(n-1-k)T_k(7,1)T_k(10,10)^2)/(n10^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

A329107 a(n) = (Sum_{k=0..n-1}(840k+197)(-1)^k2430^(n-1-k)T_k(8,1)T_k(5,-5)^2)/(n5^(n-1)), where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.