A010850 -id:A010850 - cp's OEIS frontend

A214676 A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

1, 1, 11, 1, 2, 111, 1, 2, 11, 1111, 1, 2, 3, 12, 11111, 1, 2, 3, 11, 21, 111111, 1, 2, 3, 4, 12, 22, 1111111, 1, 2, 3, 4, 11, 13, 111, 11111111, 1, 2, 3, 4, 5, 12, 21, 112, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 121, 1111111111

Offset: 1

Alois P. Heinz, Jul 25 2012

The digit set for bijective base-k numeration is {1, 2, ..., k}.

			Square array A(n,k) begins:
:         1,   1,  1,  1,  1,  1,  1,  1, ...
:        11,   2,  2,  2,  2,  2,  2,  2, ...
:       111,  11,  3,  3,  3,  3,  3,  3, ...
:      1111,  12, 11,  4,  4,  4,  4,  4, ...
:     11111,  21, 12, 11,  5,  5,  5,  5, ...
:    111111,  22, 13, 12, 11,  6,  6,  6, ...
:   1111111, 111, 21, 13, 12, 11,  7,  7, ...
:  11111111, 112, 22, 14, 13, 12, 11,  8, ...

Alois P. Heinz, Antidiagonals n = 1..18, flattened
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995, 27-29.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration

Columns k=1-9 give: A000042, A007931, A007932, A084544, A084545, A057436, A214677, A214678, A052382.

A(n+1,n) gives A010850.

A:= proc(n, b) local d, l, m; m:= n; l:= NULL;
      while m>0 do  d:= irem(m, b, 'm');
        if d=0 then d:=b; m:=m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);

A[n_, b_] := Module[{d, l, m}, m = n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
Table[A[n, d-n+1], {d, 1, 12}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A023010 Number of partitions of n into parts of 11 kinds.

Original entry on oeis.org

1, 11, 77, 418, 1925, 7854, 29183, 100529, 325193, 997150, 2919411, 8207563, 22259237, 58454165, 149104450, 370410700, 898202998, 2130141651, 4949034937, 11281187225, 25262712629, 55641782779, 120661583781, 257862888360, 543532730675, 1130864017283

Offset: 0

David W. Wilson

nonn

a(n) is Euler transform of A010850. - Alois P. Heinz, Oct 17 2008

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for expansions of Product_{k >= 1} (1-x^k)^m
N. J. A. Sloane, Transforms

Cf. 11th column of A144064. - Alois P. Heinz, Oct 17 2008

with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(add(d*11, d=divisors(j)) *a(n-j), j=1..n)/n) end: seq(a(n), n=0..40); # Alois P. Heinz, Oct 17 2008

nmax=50; CoefficientList[Series[Product[1/(1-x^k)^11,{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Feb 28 2015 *)
CoefficientList[Series[1/QPochhammer[x]^11, {x, 0, 30}], x] (* Indranil Ghosh, Mar 27 2017 *)

Vec(1/eta(x)^11 + O(x^30)) \\ Indranil Ghosh, Mar 27 2017

G.f.: Product_{m>=1} 1/(1-x^m)^11.
a(n) ~ 1331 * exp(Pi * sqrt(22*n/3)) / (2^(19/2) * 27 * n^(7/2)). - Vaclav Kotesovec, Feb 28 2015
a(0) = 1, a(n) = (11/n)*Sum_{k=1..n} A000203(k)*a(n-k) for n > 0. - Seiichi Manyama, Mar 27 2017
G.f.: exp(11*Sum_{k>=1} x^k/(k*(1 - x^k))). - Ilya Gutkovskiy, Feb 06 2018

A156194 Period 12: 1,2,7,1,7,2,1,1,4,2,4,1 repeated.

Original entry on oeis.org

1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1, 1, 2, 7, 1, 7, 2, 1, 1, 4, 2, 4, 1

Offset: 0

Paul Curtz, Feb 05 2009

Also the decimal expansion of 42390704747/333333333333. - R. J. Mathar, Feb 23 2009

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1).

PadRight[{},144,{1,2,7,1,7,2,1,1,4,2,4,1}] (* Harvey P. Dale, Mar 06 2012 *)

Palindromic properties: a(12k+i) = a(12k+6-i), i=0..3. a(12k+7+i) = a(12k+11-i), i=0..2, and similarly for successive differences.
a(n) = A156095(n) mod 9.
a(n) = A156094(n+6) mod 9.
a(4n) + a(4n+1) + a(4n+2) + a(4n+3) = A010850(n).
G.f.: (1+2*x+7*x^2+x^3+7*x^4+2*x^5+x^6+x^7+4*x^8+2*x^9+4*x^10+x^11)/((1-x)*(1+x+x^2)*(1+x)*(1-x+x^2)*(1+x^2)*(x^4-x^2+1)). - R. J. Mathar, Feb 23 2009

Edited by R. J. Mathar, Feb 23 2009

More terms from Jinyuan Wang, Feb 26 2020

A299781 Triangle read by rows: T(n,m) = number of k-uniform tilings having m different arrangements of polygons about their vertices, for k = 1..n, with 1 <= m <= n.

Original entry on oeis.org

11, 11, 20, 11, 42, 39, 11, 75, 124, 33, 11, 149, 273, 127, 15, 11, 249, 557, 314, 107, 10, 11

Offset: 1

Omar E. Pol, Mar 30 2018

Column m lists the partial sums of the m-th column of triangle A299780.

			Triangle begins:
  11;
  11,  20;
  11,  42,  39;
  11,  75, 124,  33;
  11, 149, 273, 127,  15;
  11, 249, 557, 314, 107, 10;
...

Brian L. Galebach, n-Uniform Tilings

Column 1 gives A010850.
Leading diagonal is A068600.
Row sums give A299782.
Cf. A068599, A299780.

A368782 Comma transform of A366487.

Original entry on oeis.org

12, 35, 94, 15, 16, 28, 31, 34, 37, 41, 45, 55, 55, 55, 55, 61, 67, 74, 71, 89, 98, 97, 18, 19, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 11, 11, 12, 13, 14, 15, 16, 17, 18, 22, 22, 24, 26, 28, 22, 22, 24, 26, 28, 22, 22, 24, 26, 28, 22, 22, 24, 26, 28, 22, 22

Offset: 1

Michael S. Branicky, Jan 05 2024

See A367360 for further information.
Let the comma sequence A121805 be known as S or C0.
A366487, the first differences of A121805, is the same as the comma transform of A121805; call it C1.
This sequence is C2 = C(C(S)), the comma transform C iterated twice.
C4 = C2, C5 = C2, ... once the first term (and the last term if the sequence is finite) are removed from the lower iterates of C.
Theorem: C^{i+2}(S) = C^i(S) for i>=2 in general and for i>=0 when all terms of S have two digits and no least significant digit is zero. See link for proof.
Remark. The lexicographically earliest sequence S with C(S) = S is A010850, all 11's.
The sequence contains 2137451 terms, with a(2137451) = 96. The next term does not exist.

Michael S. Branicky, Table of n, a(n) for n = 1..20000
Michael S. Branicky, Comma Comma Proof

Cf. A010850, A121805, A367360, A366487.

from itertools import islice, pairwise
def S(): # generator of comma sequence
    an = 1
    while True:
        yield an
        an += 10*(an%10)
        children = [an+y for y in range(1, 10) if str(an+y)[0] == str(y)]
        if not children: break
        an = children[0]
def C(g): # generator of comma transform of sequence passed as a generator
    yield from (10*(t%10) + int(str(u)[0]) for t, u in pairwise(g))
def agen(): return C(C(S()))
print(list(islice(agen(), 70))) # Michael S. Branicky, Jan 05 2024

A267319 Continued fraction expansion of phi^8, where phi = (1 + sqrt(5))/2.

Original entry on oeis.org

46, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1, 45, 1

Offset: 0

Ilya Gutkovskiy, Jan 13 2016

More generally, the ordinary generating function for the continued fraction expansion of phi^(2*k + 1), where phi = (1 + sqrt(5))/2, k = 1, 2, 3,... is floor(phi^(2*k + 1))/(1 - x), and for the continued fraction expansion of phi^(2*k) is (floor(phi^(2*k)) + x - x^2)/(1 - x^2).

			phi^8 = (47 + 21*sqrt(5))/2 = 46 + 1/(1 + 1/(45 + 1/(1 + 1/(45 + 1/(1 + 1/(45 + 1/...)))))).

Eric Weisstein's World of Mathematics, Golden Ratio
Wikipedia, Golden ratio
Index entries for continued fractions for constants
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A001622.

Cf. continued fraction expansion of phi^k: A000012 (k = 1), A054977 (k = 2), A010709 (k = 3), A176260 (k = 4, for n>0), A010850 (k = 5), A040071 (k = 6, for n>0), A010868 (k = 7), this sequence (k = 8).

[46] cat &cat [[1, 45]^^50]; // Vincenzo Librandi, Jan 13 2016

ContinuedFraction[(47 + 21 Sqrt[5])/2, 82]

G.f.: (46 + x - x^2)/(1 - x^2).

a(n) = 23 + 22*(-1)^n for n>0. - Bruno Berselli, Jan 18 2016

A267615 a(n) = 2^n + 11.

Original entry on oeis.org

12, 13, 15, 19, 27, 43, 75, 139, 267, 523, 1035, 2059, 4107, 8203, 16395, 32779, 65547, 131083, 262155, 524299, 1048587, 2097163, 4194315, 8388619, 16777227, 33554443, 67108875, 134217739, 268435467, 536870923, 1073741835, 2147483659, 4294967307, 8589934603, 17179869195, 34359738379

Offset: 0

Ilya Gutkovskiy, Jan 18 2016

Recurrence relation b(n) = 3*b(n - 1) - 2*b(n - 2) for n>1, b(0) = k, b(1) = k + 1, gives the closed form b(n) = 2^n + k - 1.

Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. sequences with closed form 2^n + k - 1: A168616 (k=-4), A028399 (k=-3), A036563 (k=-2), A000918 (k=-1), A000225 (k=0), A000079 (k=1), A000051 (k=2), A052548 (k=3), A062709 (k=4), A140504 (k=5), A168614 (k=6), A153972 (k=7), A168415 (k=8), A242475 (k=9), A188165 (k=10), A246139 (k=11), this sequence (k=12).

Cf. A156940.

[2^n+11: n in [0..30]]; // Vincenzo Librandi, Jan 19 2016

Table[2^n + 11, {n, 0, 35}]
LinearRecurrence[{3, -2}, {12, 13}, 40] (* Vincenzo Librandi, Jan 19 2016 *)

a(n) = 2^n + 11; \\ Altug Alkan, Jan 18 2016

G.f.: (12 - 23*x)/(1 - 3*x + 2*x^2).
a(n) = 3*a(n - 1) - 2*a(n - 2) for n>1, a(0)=12, a(1)=13.
a(n) = A000079(n) + A010850(n).
Sum_{n>=0} 1/a(n) = 0.367971714327125...
Lim_{n->oo} a(n + 1)/a(n) = 2.
E.g.f.: exp(2*x) + 11*exp(x). - Elmo R. Oliveira, Nov 08 2023

cp's OEIS Frontend

A214676 A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

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A023010 Number of partitions of n into parts of 11 kinds.

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A156194 Period 12: 1,2,7,1,7,2,1,1,4,2,4,1 repeated.

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Extensions

A299781 Triangle read by rows: T(n,m) = number of k-uniform tilings having m different arrangements of polygons about their vertices, for k = 1..n, with 1 <= m <= n.

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A368782 Comma transform of A366487.

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A267319 Continued fraction expansion of phi^8, where phi = (1 + sqrt(5))/2.

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Magma

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A267615 a(n) = 2^n + 11.

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