A167819 -id:A167819 - cp's OEIS frontend

A218566 Triangle T[r,c]=(r-1)binomial(r-1,c-1)(c-1)!*A093883(c), read by rows.

0, 1, 3, 2, 12, 240, 3, 27, 1080, 226800, 4, 48, 2880, 1209600, 3657830400, 5, 75, 6000, 3780000, 22861440000, 1267438233600000, 6, 108, 10800, 9072000, 82301184000, 9125555281920000, 11274806061917798400000

Offset: 1

M. F. Hasler, Nov 02 2012

T[b,d] gives the number of positive numbers that can be written in base b with d(d+1)/2 digits such that for each k=1,...,d some digit appears exactly k times, cf. A218560, A167819, A218556 and related sequences.

			The first 6 rows of the triangle are:
r=1: 0;
r=2: 1, 3;
r=3: 2, 12,  240;
r=4: 3, 27,  1080,  226800;
r=5: 4, 48,  2880,  1209600,  3657830400;
r=6: 5, 75,  6000,  3780000,  22861440000,  1267438233600000.
Row 2 counts the numbers 1 and 4=100[2], 5=101[2], 6=110[2].
Row 3 counts the numbers {1, 2} and {9=100[3], 10=101[3], 12=110[3], 14=112[3], 16=121[3], ..., 25=221[3]} and {248=100012[3], ..., 714=222110[3]}.

T(r,c)=(r-1)*binomial(r-1,c-1)*(c-1)!*A093883(c)

T[r,1] = r-1. T[r,2] = 3(r-1)^2. T[r,3] = 60(r-2)(r-1)^2, etc.

A218560 Numbers with d distinct ternary digits (d=1,2,3) such that for each k=1,...,d, some digit occurs exactly k times.

Original entry on oeis.org

0, 1, 2, 9, 10, 12, 14, 16, 17, 18, 20, 22, 23, 24, 25, 248, 250, 251, 254, 257, 258, 259, 262, 263, 264, 265, 267, 269, 272, 275, 276, 277, 281, 285, 287, 288, 289, 291, 293, 295, 296, 298, 299, 300, 301, 303, 305, 306, 307, 309, 311, 313, 314, 315, 317, 319, 320, 321, 322, 326, 329, 330, 331, 335

Offset: 1

M. F. Hasler, Nov 02 2012

For each of the terms, the number of ternary (= base 3) digits is a triangular number A000217.
The base 2 analog would have only the 5 terms 0,1,4,5,6. See A218556 for the base 10 analog.
The sequence A167819 is a subsequence containing exactly all terms >= 9.
The sequence is finite, with 255=3+12+240 (= 1 + sum of the 3rd row of A218566) terms.

			The terms a(1)=0 through a(3)=2 have exactly 1 digit occurring exactly once.
The terms a(4)=9=100[3] through a(15)=25=221[3], have one ternary digit occurring once and a second, different digit occurring exactly twice.
The terms a(16)=248=100012[3] through a(255)=714=222110[3] contain each ternary digit at least once. There are no other terms in this sequence.

M. F. Hasler, Table of n, a(n) for n = 1..255 (full sequence).

Cf. A218559, A182040, A218556.

{my(T(n)=n*(n+1)\2); print1(0); for(i=1,3, s=vector(i+1,j,j-1); for(n=3^(T(i)-1),3^T(i)-1,i !=#Set(digits(n,3)) & next; c=vector(4); for(j=1,#d=digits(n,3),c[d[j]+1]++); vecsort(c,,8)==s & print1(","n)))}

is_A218560(n,b=3)={ my(c=vector(b+1)); for(i=1,#n=digits(n,b),c[n[i]+1]++); #(c=vecsort(c,,8))==1+c[#c] && 2*#n==c[#c]*#c }

A218559 Sum_{i=0..n-1} i(n^(i+1)-1)/(n-1)n^(i(i+1)/2).

Original entry on oeis.org

0, 6, 714, 1047188, 30515132780, 21936856591278330, 459986443452971306412268, 324518550895166392891543292552264, 8727963565271662417355532872177263437534624, 9999999999888888888777777776666666555555444443333222110

Offset: 1

M. F. Hasler, Nov 02 2012

Largest number which can be written in base n using d+1 times the digit d, d=0,...,n-1. (Or: such that for each k=1,...,n, some digit is used exactly k times.)

			Written in the respective bases, a(2) = 6 = 110[2], a(3) = 714 = 222110[3], a(4) = 1047188 = 33322110[4], etc.

Cf. A218556, A210666, A182040; A167819, A031948.

a(b)=sum(i=1,b-1,(b^(i+1)-1)\(b-1)*b^(i*(i+1)\2)*i)

cp's OEIS Frontend

A218566 Triangle T[r,c]=(r-1)binomial(r-1,c-1)(c-1)!*A093883(c), read by rows.

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A218560 Numbers with d distinct ternary digits (d=1,2,3) such that for each k=1,...,d, some digit occurs exactly k times.

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A218559 Sum_{i=0..n-1} i(n^(i+1)-1)/(n-1)n^(i(i+1)/2).

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cp's OEIS Frontend

A218566 Triangle T[r,c]=(r-1)*binomial(r-1,c-1)*(c-1)!*A093883(c), read by rows.

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A218560 Numbers with d distinct ternary digits (d=1,2,3) such that for each k=1,...,d, some digit occurs exactly k times.

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PARI

PARI

A218559 Sum_{i=0..n-1} i*(n^(i+1)-1)/(n-1)*n^(i(i+1)/2).

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A218566 Triangle T[r,c]=(r-1)binomial(r-1,c-1)(c-1)!*A093883(c), read by rows.

A218559 Sum_{i=0..n-1} i(n^(i+1)-1)/(n-1)n^(i(i+1)/2).