A329107 -id:A329107 - cp's OEIS frontend

A328786 a(n) = (Sum_{k=0..n-1}(-1)^k(39480k+7321)29700^(n-1-k)T_k(14,1)T_k(11,-11)^2)/n, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

7321, 69076403, 1423525024746, 31676475535509475, 752633551945067097470, 18627509719518121995003486, 474204125641606160260805604468, 12323377272130975561953028453412931, 325337163371714764552775702345400136950, 8696262375383068237957170325229215635055690

Offset: 1

Zhi-Wei Sun, Nov 05 2019

nonn

Conjecture: (i) a(n) is a positive integer for each n > 0, and a(n) is odd if and only if n is a power of two. Moreover, we have the identity Sum_{k>=0}(39480*k+7321)(-29700)^(-k)*T_k(14,1)*T_k(11,-11)^2 = 6795*sqrt(5)/Pi.
(ii) For any prime p > 5, we have the congruence Sum_{k=0..p-1}(39480*k+7321)(-29700)^(-k)*T_k(14,1)*T_k(11,-11)^2 == p*(1513 + 70*Leg(3/p) + 5738*Leg(-5/p)) (mod p^2), where Leg(a/p) denotes the Legendre symbol.
(iii) Let p > 5 be a prime different from 11, and set S(p) = Sum_{k=0..p-1} T_k(14,1)*T_k(11,-11)^2/(-29700)^k. If Leg(-165/p) = -1, then S(p) == 0 (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/11) = 1 and p = x^2 + 165*y^2 (with x and y integers), then S(p) == 4*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = Leg(p/5) = Leg(p/11) = -1 and 2p = x^2 + 165*y^2, then S(p) == 2*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = -1, Leg(p/3) = Leg(p/11) = 1, and p = 3*x^2 + 55*y^2, then S(p) == 12*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/5) = 1, Leg(p/3) = Leg(p/11) = -1, and 2p = 3*x^2 + 55*y^2, then S(p) == 6*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/11) = 1, Leg(p/3) = Leg(p/5) = -1, and p = 5*x^2 + 33*y^2, then S(p) == 2p-20*x^2 (mod p^2). If Leg(-1/p) = Leg(p/11) = -1, Leg(p/3) = Leg(p/5) = 1, and 2p = 5*x^2 + 33*y^2, then S(p) == 2p-10*x^2 (mod p^2). If Leg(-1/p) = Leg(p/3) = -1, Leg(p/5) = Leg(p/11) = 1, and p = 11*x^2 + 15*y^2, then S(p) == 44*x^2-2p (mod p^2). If Leg(-1/p) = Leg(p/3) = 1, Leg(p/5) = Leg(p/11) = -1, and 2p = 11*x^2 + 15*y^2, then S(p) == 22*x^2-2p (mod p^2).
In 2011 the author formulated many conjectural series for 1/Pi involving T_k(b,c). The conjectural series for 1/Pi in A329073, A329105, A329107 and the present one are of a new type which might be called type VIII to distinguish from the 7 known types (types I-VII).

			a(1) = 7321 since ((-1)^0*(39480*0+7321)*29700^(1-1-0)*T_0(14,1)*T_0(11,-11)^2)/1 = 7321.

Zhi-Wei Sun, Table of n, a(n) for n = 1..40
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. A329073, A329105, 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[(39480k+7321)T[14,1,k]T[11,-11,k]^2*(-1)^k*29700^(n-1-k),{k,0,n-1}]/n;
Table[a[n],{n,1,10}]

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.

Original entry on oeis.org

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}]

cp's OEIS Frontend

A328786 a(n) = (Sum_{k=0..n-1}(-1)^k(39480k+7321)29700^(n-1-k)T_k(14,1)T_k(11,-11)^2)/n, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

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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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cp's OEIS Frontend

A328786 a(n) = (Sum_{k=0..n-1}(-1)^k*(39480*k+7321)*29700^(n-1-k)*T_k(14,1)*T_k(11,-11)^2)/n, 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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Author

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Comments

Examples

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Mathematica

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A328786 a(n) = (Sum_{k=0..n-1}(-1)^k(39480k+7321)29700^(n-1-k)T_k(14,1)T_k(11,-11)^2)/n, where T_k(b,c) denotes the coefficient of x^k in the expansion of (x^2+bx+c)^k.

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.