A101121 -id:A101121 - cp's OEIS frontend

A101120 Records in A101119, which forms the nonzero differences of A006519 and A003484.

7, 22, 52, 112, 239, 494, 1004, 2024, 4071, 8166, 16356, 32736, 65503, 131038, 262108, 524248, 1048535, 2097110, 4194260, 8388560, 16777167, 33554382, 67108812, 134217672, 268435399, 536870854, 1073741764, 2147483584, 4294967231, 8589934526, 17179869116, 34359738296

Offset: 1

Simon Plouffe and Paul D. Hanna, Dec 02 2004

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,0,1,-3,2).

Cf. A003484, A006519, A101119, A101121, A101122.

LinearRecurrence[{3,-2,0,1,-3,2},{7,22,52,112,239,494},30] (* Harvey P. Dale, Jan 23 2023 *)

PARI
```
a(n)=2^(n+3)-2^((n-1)%4)-8*((n+3)\4)
```

def A101120(n): return (1<<(n+3))-(1<<((n-1)&3))-(((n+3)&-4)<<1) # Chai Wah Wu, Jul 10 2022

a(n) = A101119(2^(n-1)) for n>=1.
a(n) = 2^(n+3) - 2^((n-1)(mod 4)) - 8*floor((n+3)/4).
a(n) = 2^(n+3) - A003485(n+3). - Johannes W. Meijer, Oct 31 2012
From Chai Wah Wu, Apr 15 2017: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) + a(n-4) - 3*a(n-5) + 2*a(n-6) for n > 6.
G.f.: x*(-x - 7)/((x - 1)^2*(x + 1)*(2*x - 1)*(x^2 + 1)). (End)
E.g.f.: (exp(x)*(32*exp(x) - 8*x - 27) - 4*cos(x) - cosh(x) - 2*sin(x) + sinh(x))/4. - Stefano Spezia, Jun 06 2023

A101122 XOR BINOMIAL transform of A101119.

Original entry on oeis.org

7, 17, 0, 34, 0, 0, 0, 68, 0, 0, 0, 0, 0, 0, 0, 159, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 257, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 514, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Simon Plouffe and Paul D. Hanna, Dec 02 2004

nonn

Nonzero terms form A101121 and occur at positions 2^k for k >= 0. A101119 equals the nonzero differences of A006519 and A003484. See A099884 for the definition of the XOR BINOMIAL transform.

Cf. A003484, A006519, A101119, A101120, A101121.

{a(n)=local(B);B=0;for(i=0,n-1,B=bitxor(B,binomial(n-1,i)%2* (16*2^valuation(n-i,2)-2^(valuation(n-i,2)%4)-8*(valuation(n-i,2)\4)-8)));B}

from operator import xor
from functools import reduce
def A101122(n): return reduce(xor,(((1<<(m:=(~(k+1)&k).bit_length()+4))-((m&-4)<<1)-(1<<(m&3)))&-int(not k&~(n-1)) for k in range(n))) # Chai Wah Wu, Jul 10 2022

a(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*A101119(k), where SumXOR is summation under XOR. A101119(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*a(k). a(2^(n-1)) = A101121(n) for n >= 1 and a(k)=0 when k is not a power of 2.

cp's OEIS Frontend

A101120 Records in A101119, which forms the nonzero differences of A006519 and A003484.

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