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 1-3 of 3 results.

A099902 Multiplies by 2 and shifts right under the XOR BINOMIAL transform (A099901).

Original entry on oeis.org

1, 3, 7, 11, 23, 59, 103, 139, 279, 827, 1895, 2955, 5655, 14395, 24679, 32907, 65815, 197435, 460647, 723851, 1512983, 3881019, 6774887, 9142411, 18219287, 54002491, 123733863, 192940939, 369104407, 939538491, 1610637415, 2147516555
Offset: 0

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Author

Paul D. Hanna, Oct 30 2004

Keywords

Comments

Equals the XOR BINOMIAL transform of A099901. Also, equals the main diagonal of the XOR difference triangle A099900, in which the central terms of the rows form the powers of 2.
Bisection of A101624. - Paul Barry, May 10 2005

Crossrefs

Programs

  • Maple
    a:= n -> add((binomial(n-k+floor(k/2),floor(k/2)) mod 2)*2^k, k=0..n):
    map(a, [$0..100]); # Robert Israel, Jan 24 2016
  • PARI
    {a(n)=local(B);B=0;for(k=0,n,B=bitxor(B,binomial(n-k+k\2,k\2)%2*2^k));B}
    
  • PARI
    a(n)=sum(k=0,n,binomial(n-k+k\2,k\2)%2*2^k)
    
  • Python
    def A099902(n): return sum(int(not ~((n<<1)-k)&k)<Chai Wah Wu, Jul 30 2025

Formula

a(n) = SumXOR_{k=0..n} (binomial(n-k+floor(k/2), floor(k/2)) mod 2)*2^k for n >= 0.
a(n) = SumXOR_{i=0..n} (C(n, i) mod 2)*A099901(n-i), where SumXOR is the analog of summation under the binary XOR operation and C(i, j) mod 2 = A047999(i, j).
a(n) = Sum_{k=0..n} A047999(n-k+floor(k/2), floor(k/2)) * 2^k.
From Paul Barry, May 10 2005: (Start)
a(n) = Sum_{k=0..2n} (binomial(k, 2n-k) mod 2)*2^(2n-k);
a(n) = Sum_{k=0..n} (binomial(2n-k, k) mod 2)*2^k. (End)
a(n) = Sum_{k=0..2n} A106344(2n,k)*2^(2n-k). - Philippe Deléham, Dec 18 2008

A099897 XOR difference triangle, read by rows, of A099898 (in leftmost column) such that the main diagonal equals A099898 shift left and divided by 4.

Original entry on oeis.org

1, 4, 5, 20, 16, 21, 84, 64, 80, 69, 276, 320, 256, 336, 277, 1108, 1344, 1024, 1280, 1104, 1349, 5396, 4416, 5120, 4096, 5376, 4432, 5141, 20564, 17728, 21504, 16384, 20480, 17664, 21584, 16453, 65812, 86336, 70656, 81920, 65536, 86016, 70912
Offset: 0

Views

Author

Paul D. Hanna, Oct 30 2004

Keywords

Comments

Central terms of rows equal powers of 4: T(n,[n/2]) = 4^n for n>=0. The leftmost column is A099898. The diagonal forms A099899 and equals the XOR BINOMIAL transform of A099898. See A099884 for the definitions of XOR difference triangle and the XOR BINOMIAL transform.

Examples

			Rows begin:
[_1],
[_4,5],
[20,_16,21],
[84,_64,80,69],
[276,320,_256,336,277],
[1108,1344,_1024,1280,1104,1349],
[5396,4416,5120,_4096,5376,4432,5141],
[20564,17728,21504,_16384,20480,17664,21584,16453],
[65812,86336,70656,81920,_65536,86016,70912,82256,65813],...
notice that the column terms equal 4 times the diagonal (with offset), and that the central terms in the rows form the powers of 4.
		

Crossrefs

Programs

  • PARI
    T(n,k)=if(n
    				

Formula

T(n, [n/2]) = 4^n. T(n+1, 0) = 4*T(n, n) (n>=0); T(0, 0)=1; T(n, k) = T(n, k-1) XOR T(n-1, k-1) for n>k>0. T(n, k) = SumXOR_{i=0..k} (C(k, i)mod 2)*T(n-i, 0), where SumXOR is the analog of summation under the binary XOR operation and C(k, i)mod 2 = A047999(k, i).

A099901 Shifts left and divides by 2 under the XOR BINOMIAL transform (A099902).

Original entry on oeis.org

1, 2, 6, 14, 22, 46, 118, 206, 278, 558, 1654, 3790, 5910, 11310, 28790, 49358, 65814, 131630, 394870, 921294, 1447702, 3025966, 7762038, 13549774, 18284822, 36438574, 108004982, 247467726, 385881878, 738208814, 1879076982, 3221274830
Offset: 0

Views

Author

Paul D. Hanna, Oct 30 2004

Keywords

Comments

Equals the XOR BINOMIAL transform of A099902. Also, equals the leftmost column of the XOR difference triangle A099900, in which the central terms of the rows forms the powers of 2.

Crossrefs

Programs

  • PARI
    {a(n)=local(B);B=0;if(n==0,B=1, for(k=0,n-1, B=bitxor(B,binomial(n-1-k+k\2,k\2)%2*2^(k+1))));B}

Formula

a(0)=1; a(n) = SumXOR_{k=0..n-1} (C(n-1-k+[k/2], [k/2])mod 2)*2^(k+1) for n>0. a(n) = SumXOR_{i=0..n} (C(n, i)mod 2)*A099902(n-i), where SumXOR is the analog of summation under the binary XOR operation and C(k, i)mod 2 = A047999(k, i).
Showing 1-3 of 3 results.