A038192 -id:A038192 - cp's OEIS frontend

A089893 a(n) = (A001317(2n)-1)/4.

0, 1, 4, 21, 64, 321, 1092, 5461, 16384, 81921, 278532, 1392661, 4210752, 21053761, 71582788, 357913941, 1073741824, 5368709121, 18253611012, 91268055061, 275951648832, 1379758244161, 4691178030148, 23455890150741

Offset: 0

Ralf Stephan, Jan 10 2004

nonn

a(n) = (A038192(n) - 3)/12 = (A038183(n) - 1)/4, reflecting the fact that, if you look at each row of the Pascal triangle's parity as a binary number (cf. A001317), then the numbers in odd rows are thrice the numbers in even rows.
Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true. - Vladimir Shevelev, Nov 28 2010]
Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - Vladimir Shevelev, Nov 28 2010]

Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, arXiv:1011.6083 [math.NT], 2010-2012.

Cf. A000215, A001317, A038183, A038192.

a1317[n_] := Sum[2^k Mod[Binomial[n, k], 2] , {k, 0, n}];
a[n_] := (a1317[2n] - 1)/4;
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jan 18 2019 *)

a(n)=(sum(k=0,2*n+1,(binomial(2*n+1,k)%2)*2^k)-3)/12

def A089893(n): return sum((bool(~(m:=n<<1)&m-k)^1)<>2 # Chai Wah Wu, May 02 2023

A178751 Numbers k such that in Z/kZ the equation x^y + 1 = 0 has only the trivial solutions with x == -1 (mod k).

Original entry on oeis.org

2, 3, 4, 6, 8, 12, 15, 16, 20, 24, 30, 32, 40, 48, 51, 60, 64, 68, 80, 96, 102, 120, 128, 136, 160, 192, 204, 240, 255, 256, 272, 320, 340, 384, 408, 480, 510, 512, 544, 640, 680, 768, 771, 816, 960, 1020, 1024, 1028, 1088, 1280, 1360, 1536, 1542, 1632, 1920, 2040

Offset: 1

Arnaud Vernier, Jun 09 2010, Jun 10 2010

nonn

It appears that odd terms 3, 15, 51, 255, 771, 3855, 13107, 65535, ... are given by A038192. - Michel Marcus, Aug 08 2013

This is the complement of A126949 in the numbers k > 1. (But it could be argued that the sequence should start with k = 1 as initial term.) It appears that for any a(j) in the sequence, 2*a(j) is also in the sequence. The primitive terms (not of the form a(j) = 2*a(m), m < j) are 2, 3, 15, 20, 51, 68, 255, 340, 771, 1028, .... (see A274003). - M. F. Hasler, Jun 06 2016

			In Z/3Z, the only solution to the equation x^y + 1 = 0 is x = 2 and y = 1. Whereas in Z/5Z, the equation has at least one nontrivial solution: 2^2 + 1 = 0.

Arnaud Vernier and Charles R Greathouse IV, Table of n, a(n) for n = 1..189 (first 78 terms from Vernier)

Cf. A038192, A126949, A274003.

is(n)=for(x=2,n-2,if(gcd(x,n)>1,next);my(t=Mod(x,n));while(abs(centerlift(t))>1,t*=x);if(t==-1,return(0)));n>1 \\ Charles R Greathouse IV, Aug 08 2013

A274003 Primitive terms (not equal twice a smaller term) of A178751: moduli n such that x^y == -1 (mod n) only for x = -1 (mod n).

Original entry on oeis.org

2, 3, 15, 20, 51, 68, 255, 340, 771, 1028, 3855, 5140, 13107, 17476, 65535, 87380, 196611, 262148, 983055

Offset: 1

M. F. Hasler, Jun 06 2016

See A178751 for further information. In particular, A038192 is the subsequence of odd terms.

Cf. A038192, A178751.

is_A274003(n)={is_A178751(n) && (bittest(n,0) || !is_A178751(n\2))}

select( n -> bittest(n,0) || !setsearch(A178751,n\2), A178751) \\ assuming the vector A178751 holds enough terms of that sequence.

cp's OEIS Frontend

A089893 a(n) = (A001317(2n)-1)/4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Python

A178751 Numbers k such that in Z/kZ the equation x^y + 1 = 0 has only the trivial solutions with x == -1 (mod k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A274003 Primitive terms (not equal twice a smaller term) of A178751: moduli n such that x^y == -1 (mod n) only for x = -1 (mod n).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

PARI