0, 1, 0, 2, 2, 0, 0, 3, 3, 3, 0, 0, 0, 0, 0, 4, 4, 4, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 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, 6, 6, 6, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0
A210792
Triangle of coefficients of polynomials v(n,x) jointly generated with A210791; see the Formula section.
Original entry on oeis.org
1, 1, 2, 1, 5, 3, 1, 10, 11, 5, 1, 19, 28, 25, 8, 1, 36, 62, 81, 50, 13, 1, 69, 129, 218, 193, 98, 21, 1, 134, 261, 533, 597, 442, 185, 34, 1, 263, 522, 1235, 1631, 1559, 952, 343, 55, 1, 520, 1040, 2773, 4129, 4763, 3758, 1985, 625, 89, 1, 1033, 2071
Offset: 1
First five rows:
1;
1, 2;
1, 5, 3;
1, 10, 11, 5;
1, 19, 28, 25, 8;
First three polynomials v(n,x):
1
1 + 2x
1 + 5x + 3x^2
From _Philippe Deléham_, Mar 29 2012: (Start)
(1, 0, 1/2, 3/2, 0, 0, 0, ...) DELTA (0, 2, -1/2, -1/2, 0, 0, 0, ...) begins:
1;
1, 0;
1, 2, 0;
1, 5, 3, 0;
1, 10, 11, 5, 0;
1, 19, 28, 25, 8, 0;
1, 36, 62, 81, 50, 13, 0;
1, 69, 129, 218, 193, 98, 21, 0; (End)
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u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + j)*v[n - 1, x] + c;
d[x_] := h + x; e[x_] := p + x;
v[n_, x_] := d[x]*u[n - 1, x] + e[x]*v[n - 1, x] + f;
j = 0; c = 0; h = -1; p = 2; f = 0;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A210791 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A210792 *)
Table[u[n, x] /. x -> 1, {n, 1, z}] (* A007051 *)
Table[v[n, x] /. x -> 1, {n, 1, z}] (* A000244 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A001129 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A001333 *)
Original entry on oeis.org
1, 2, 5, 10, 19, 36, 69, 134, 263, 520, 1033, 2058, 4107, 8204, 16397, 32782, 65551, 131088, 262161, 524306, 1048595, 2097172, 4194325, 8388630, 16777239, 33554456, 67108889, 134217754, 268435483, 536870940, 1073741853, 2147483678
Offset: 0
a(4) = 19 = sum of row 4 terms of triangle A132735: (1 + 5 + 7 + 5 + 1).
a(3) = 10 = (1, 3, 3, 1) dot (1, 1, 2, 0) = (1 + 3 + 6 + 0).
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[1] cat [2^n + n-1: n in [1..30]]; // G. C. Greubel, Feb 14 2021
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a:= proc(n) option remember; if n=0 then 1 else add((binomial(n,j)+1), j=0..n-1) fi end: seq(a(n), n=0..31);# Zerinvary Lajos, Mar 29 2009
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Table[2^n + n-1 + Boole[n==0], {n,0,30}] (* G. C. Greubel, Feb 14 2021 *)
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[1]+[2^n + n-1 for n in (1..30)] # G. C. Greubel, Feb 14 2021
A247241
In base 2, smallest positive integer k such that k contains all possible strings of n binary digits.
Original entry on oeis.org
2, 19, 558, 272060, 34949682936, 297564712194085870576, 10932581082490851526951746612437309648864, 7425338450010038712278392695279246718801455644988538421150211866664645481447360
Offset: 1
n = 1: a(1) = 2 = 10 in base 2, which contains 2 distinct strings of one digit: 1, 0.
n = 2: a(2) = 19 = 10011, which contains 4 distinct strings of two digits: 10, 00, 01, 11.
n = 3: a(3) = 558 = 1000101110, which contains 8 distinct strings of three digits: 100, 000, 001, 010, 101, 011, 111, 110.
n = 4: a(4) = 272060 = 1000010011010111100, which contains 16 distinct strings of four digits: 1000, 0000, 0001, 0010, 0100, 1001, 0011, 0110, 1101, 1010, 0101, 1011, 0111, 1111, 1110, 1100.
A342298
a(n) is the greatest k such that A342263(k) = n.
Original entry on oeis.org
2, 25, 931, 504455, 67539587599, 585462196329239562271, 21690980800898420269408456526391711768639, 14792097944732868603877386771665610972834204784426907551800717772696470224928895
Offset: 0
A309833
Maximum length of a binary sequence S such that any pair of distinct subwords of length n in S is of Hamming distance at least 3.
Original entry on oeis.org
4, 5, 7, 12, 20, 25, 39, 71
Offset: 3
A373895
a(n) = ceiling((2^n+n-1)/n).
Original entry on oeis.org
2, 3, 4, 5, 8, 12, 20, 33, 58, 104, 188, 343, 632, 1172, 2186, 4097, 7712, 14565, 27596, 52430, 99866, 190652, 364724, 699052, 1342179, 2581112, 4971028, 9586982, 18512792, 35791396, 69273668, 134217729, 260301050, 505290272, 981706812, 1908874355, 3714566312, 7233629132, 14096302922, 27487790696
Offset: 1
- Michael De Vlieger, Table of n, a(n) for n = 1..3333
- G. Ebert, J. Hammenter, F. Lazebnik, and A. Woldar, Irregularity Strengths for Certain Graphs, Congressus Numerantium 71 (1990), 39-52.
- Eric Weisstein's World of Mathematics, Hypercube Graph.
- Eric Weisstein's World of Mathematics, Irregularity Strength.
A091264
Matrix defined by a(n,k) = 2^n + (k-1), read by antidiagonals.
Original entry on oeis.org
0, 1, 1, 3, 2, 2, 7, 4, 3, 3, 15, 8, 5, 4, 4, 31, 16, 9, 6, 5, 5, 63, 32, 17, 10, 7, 6, 6, 127, 64, 33, 18, 11, 8, 7, 7, 255, 128, 65, 34, 19, 12, 9, 8, 8, 511, 256, 129, 66, 35, 20, 13, 10, 9, 9, 1023, 512, 257, 130, 67, 36, 21, 14, 11, 10, 10, 2047, 1024, 513, 258, 131, 68, 37, 22
Offset: 0
{0};
{1,1};
{3,2,2};
{7,4,3,3};
{15,8,5,4,4};
{31,16,9,6,5,5};
{63,32,17,10,7,6,6};
a(5,3) = 34 because 2^5 + (3-1) = 34.
Rows: a(0, k) =
A001477(k), a(1, k) =
A000027(k+1) etc. etc. Columns: a(n, 0) =
A000225(n). a(n, 1) =
A000079(n). a(n, 2) =
A000051(n). a(n, 3) =
A052548(n). a(n, 4) =
A062709(n). Diagonals: a(n, n+3) =
A052968(n+1). a(n, n+2) =
A005126(n). a(n, n+1) =
A006127(n). a(n, n) =
A052944(n). a(n, n-1) =
A083706(n-1). Also note that the sums of the antidiagonals = the partial sums of the main diagonal, i.e., a(n, n).
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Flatten[ Table[ Table[ a[i, n - i], {i, n, 0, -1}], {n, 0, 11}]] (* both from Robert G. Wilson v, Feb 26 2004 *)
Table[a[n, k], {n, 0, 10}, {k, 0, 10}] // TableForm (* to view the table *)
Comments