cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A178747 Sum of terms in 'rows' of A178746.

Original entry on oeis.org

1, 3, 19, 65, 295, 1129, 4663, 18441, 74359, 296585, 1188727, 4751497, 19015543, 76048521, 304232311, 1216874633, 4867651447, 19470387337, 77882161015, 311527770249, 1246113527671, 4984450615433, 19937812248439, 79751235012745, 319004979197815, 1276019860867209
Offset: 0

Views

Author

David Scambler, Jun 09 2010

Keywords

Examples

			a(0) = 1, a(1) = 3, a(2) = 6 + 6 + 7 = 19.

Links

Crossrefs

Cf. A178748 (sum of '1' bits in rows of A178746).
Cf. A001045, A081254, A138238.

Programs

  • PARI
    seq(n)={my(a=vector(n+1), f=0, p=0, k=1, s=0); while(k<=#a, my(b=bitxor(p+1,p)); f=bitxor(f,b); p=bitxor(p, bitand(b,f)); if(p>2^k, a[k]=s; k++; s=0); s+=p); a} \\ Andrew Howroyd, Mar 03 2020
  • PARI
    a(n) = {(17*4^n + 5*(2*(-1)^n-1)*2^n - 7*(-1)^n)/15} \\ Andrew Howroyd, Mar 03 2020

Formula

G.f: (1/4)*x^3 - (1/8)*x^2 - 1/16 + (x^4 + (3/4)*x^3 - (1/2)*x^2 - (3/16)*x + 1/16)*F(x) = 0. [From GUESSS]
From David Scambler, Jun 17 2010: (Start)
a(n) = (17*4^n + 5*(2*(-1)^n-1)*2^n - 7*(-1)^n)/15.
a(n) = A001045(n+1) * A081254(n+1) + (-1)^n * A138238(n-1).
(End)

Extensions

Terms a(16) and beyond from Andrew Howroyd, Mar 03 2020

A178748 Total number of '1' bits in the terms of 'rows' of A178746.

Original entry on oeis.org

1, 2, 7, 14, 37, 80, 187, 410, 913, 1988, 4327, 9326, 20029, 42776, 91027, 192962, 407785, 859244, 1805887, 3786518, 7922581, 16544192, 34486507, 71769194, 149130817, 309446420, 641262487, 1327264190, 2744006893, 5666970728, 11691855427, 24099538706, 49630733209
Offset: 0

Views

Author

David Scambler, Jun 09 2010

Keywords

Comments

Sum of adjacent terms equals the difference of adjacent terms in A127981. - David Scambler, Jun 10 2010

Examples

			a(0) = bitcount(1) = 1.
a(1) = bitcount(3) = 2.
a(2) = bitcount(6) + bitcount(6) + bitcount(7) = 2 + 2 + 3 = 7.

Links

Crossrefs

Cf. A178747 (sum of terms in rows of A178746).
Cf. A127981.

Programs

  • Mathematica
    LinearRecurrence[{2,3,-4,-4},{1,2,7,14},40] (* Harvey P. Dale, Aug 27 2021 *)
  • PARI
    seq(n)={my(a=vector(n+1), f=0, p=0, k=1, s=0); while(k<=#a, my(b=bitxor(p+1,p)); f=bitxor(f,b); p=bitxor(p, bitand(b,f)); if(p>2^k, a[k]=s; k++; s=0); s+=hammingweight(p)); a} \\ Andrew Howroyd, Mar 03 2020
  • PARI
    a(n) = {(2^n*(3*n+8) + (3*n+1)*(-1)^n)/9} \\ Andrew Howroyd, Mar 03 2020
  • PARI
    Vec((1 - 2*x^3) / ((1 + x)^2*(1 - 2*x)^2) + O(x^30)) \\ Colin Barker, Mar 04 2020

Formula

G.f: (1/2)*x^3 - 1/4 + (x^4 + x^3 - (3/4)*x^2 - (1/2)*x + 1/4)*F(x) = 0. [From GUESSS]
From David Scambler, Jun 10 2010: (Start)
a(n) = (2^n*(3*n+8) + (3*n+1)*(-1)^n)/9.
a(n) + a(n-1) = A127981(n+1) - A127981(n).
(End)
From Colin Barker, Mar 04 2020: (Start)
G.f.: (1 - 2*x^3) / ((1 + x)^2*(1 - 2*x)^2).
a(n) = 2*a(n-1) + 3*a(n-2) - 4*a(n-3) - 4*a(n-4) for n>3.
(End)

Extensions

Terms a(16) and beyond from Andrew Howroyd, Mar 03 2020
Showing 1-2 of 2 results.

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