A141725 -id:A141725 - cp's OEIS frontend

A344920 The Worpitzky transform of the squares.

0, -1, 5, -13, 29, -61, 125, -253, 509, -1021, 2045, -4093, 8189, -16381, 32765, -65533, 131069, -262141, 524285, -1048573, 2097149, -4194301, 8388605, -16777213, 33554429, -67108861, 134217725, -268435453, 536870909, -1073741821, 2147483645, -4294967293

Offset: 0

Peter Luschny, Jun 24 2021

The Worpitzky transform maps a sequence A to a sequence B, where B(n) = Sum_{k=0..n} A163626(n, k)*A(k). (If A(n) = 1/(n + 1) then B(n) are the Bernoulli numbers (with B(1) = 1/2.))

Also row 2 in A371761. Can be generated by the signed Akiyama-Tanigawa algorithm for powers (see the Python script). - Peter Luschny, Apr 12 2024

Index entries for linear recurrences with constant coefficients, signature (-3,-2).

Up to shift and sign: even bisection A267921, odd bisection A141725.

Cf. A163626, A036563, A371761.

gf := (exp(x) - 1)*(exp(x) - 2)*exp(-2*x): ser := series(gf, x, 36):
seq(n!*coeff(ser, x, n), n = 0..31);

W[n_, k_] := (-1)^k k! StirlingS2[n + 1, k + 1];
WT[a_, len_] := Table[Sum[W[n, k] a[k], {k, 0, n}], {n, 0, len-1}];
WT[#^2 &, 32] (* The Worpitzky transform applied to the squares. *)

# Using the Akiyama-Tanigawa algorithm for powers from A371761.
print([(-1)**n * v for (n, v) in enumerate(ATPowList(2, 32))])
# Peter Luschny, Apr 12 2024

a(n) = n! * [x^n] (exp(x) - 1)*(exp(x) - 2)*exp(-2*x).
a(n) = (-1)^(n + 1)*(3 - 2^(n + 1)) for n >= 1. - Hugo Pfoertner, Jun 24 2021
a(n) = [x^n] x*(2*x - 1)/(2*x^2 + 3*x + 1). - Stefano Spezia, Jun 24 2021

A237419 4^(n+1)^2 - 3.

Original entry on oeis.org

1, 253, 262141, 4294967293, 1125899906842621, 4722366482869645213693, 316912650057057350374175801341, 340282366920938463463374607431768211453, 5846006549323611672814739330865132078623730171901

Offset: 0

Vincenzo Librandi, Feb 22 2014

nonn

There are no primes of this form.
Subsequence of A141725.
From Michel Lagneau, Feb 24 2014: (Start)
If n == 1, 2, 6, 7 (mod 10) then a(n) is divisible by 11.
Proof: (n+1)^2 == 4, 9 (mod 10) and the property 4^(10k) == 1 (mod 11) gives  4^(4+10k) = 4^4*4^(10k) = 256*4^(10k) == 3*1 (mod 11) and 4^(9+10k) = 4^9*4^(10k) = 262144*4^(10k) == 3*1 (mod 11). Hence 4^(n+1)^2 - 3 == 0 (mod 11). (End)
From Michel Lagneau, Feb 25 2014: (Start)
Generalization:
If n == r1,r2,r3,r4  (mod p-1) then a(n) is divisible by p if (n+1)^2 == u, v (mod p-1) with the property 4^u and 4^v == 3 mod p. (It is possible to find u = v, for example with p = 37.)
The sequence of the primes p having this property is {11, 23, 37, 59, 83, 263, 359, 383, 467, 479, 503, 563, 587, 839, 853, 887, 983, 1019, 1187, 1223, 1319, 1523, 1823, 1871, 2027, 2039, 2063, ...}.
Examples:
If n == 1,8,12,19  (mod 22) then a(n) is divisible by 23
If n == 6,10,24,28 (mod 36) then a(n) is divisible by 37
If n == 4,23,33,52 (mod 58) then a(n) is divisible by 59. (End)

			a(0) = 4^(0+1)^2 - 3 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..40

Cf. A141725.

Magma
```
[4^(n+1)^2-3: n in [0..10]];
```

A237419:=n->4^(n + 1)^2 - 3; seq(A237419(n), n=0..20); # Wesley Ivan Hurt, Feb 27 2014

Mathematica
```
Table[(4^(n + 1)^2 - 3), {n, 0, 20}]
```

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A344920 The Worpitzky transform of the squares.

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