A370369
a(n) = n! + Sum_{k=1..n-1} (n-k)*k! = n! + A014145(n-1); for n >= 2, number of m such that any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order.
Original entry on oeis.org
1, 3, 10, 37, 166, 919, 6112, 47305, 416098, 4091131, 44417044, 527456557, 6798432070, 94499679583, 1408924024696, 22425642181009, 379514672913322, 6804212771165635, 128827325000617948, 2568509718703606261, 53787877376348226574, 1180349932648067726887, 27086018941198865627200
Offset: 1
a(3) = 10 because such numbers are 0_3, 1_3, 2_3, 10_3, 12_3, 21_3, 102_3, 120_3, 201_3 and 210_3.
a(10) = 4091131 is the number of terms of A215014.
A288777
Triangle read by rows in which column k lists the positive multiples of the factorial of k, with 1 <= k <= n.
Original entry on oeis.org
1, 2, 2, 3, 4, 6, 4, 6, 12, 24, 5, 8, 18, 48, 120, 6, 10, 24, 72, 240, 720, 7, 12, 30, 96, 360, 1440, 5040, 8, 14, 36, 120, 480, 2160, 10080, 40320, 9, 16, 42, 144, 600, 2880, 15120, 80640, 362880, 10, 18, 48, 168, 720, 3600, 20160, 120960, 725760, 3628800, 11, 20, 54, 192, 840, 4320, 25200, 161280, 1088640
Offset: 1
Triangle begins:
1;
2, 2;
3, 4, 6;
4, 6, 12, 24;
5, 8, 18, 48, 120;
6, 10, 24, 72, 240, 720;
7, 12, 30, 96, 360, 1440, 5040;
8, 14, 36, 120, 480, 2160, 10080, 40320;
9, 16, 42, 144, 600, 2880, 15120, 80640, 362880;
10, 18, 48, 168, 720, 3600, 20160, 120960, 725760, 3628800;
11, 20, 54, 192, 840, 4320, 25200, 161280, 1088640, 7257600, 39916800;
...
For n = 9 and k = 2: T(9,2) is the number of numbers with two digits in A288528.
For n = 9 the row sum is 9 + 16 + 42 + 144 + 600 + 2880 + 15120 + 80640 + 362880 = 462331, the same as A014145(9) and also the same as the number of terms in A288528.
Middle diagonal gives
A001563, n>=1.
A094344
Triangle T(n,k), 0<= k <= n, read by rows; given by [0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, ...] DELTA [1, 0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] where DELTA is the operator defined in A084938.
Original entry on oeis.org
1, 0, 1, 0, 1, 1, 0, 2, 4, 1, 0, 6, 18, 13, 1, 0, 24, 96, 119, 46, 1, 0, 120, 600, 1059, 777, 199, 1, 0, 720, 4320, 9890, 10760, 5536, 1072, 1, 0, 5040, 35280, 99158, 142990, 111316, 44228, 6985, 1, 0, 40320, 322560, 1073692, 1926312, 2009578, 1217352, 395865, 53218, 1
Offset: 0
Triangle begins:
1;
0, 1;
0, 1, 1;
0, 2, 4, 1;
0, 6, 18, 13, 1;
0, 24, 96, 119, 46, 1;
...
A200545
Triangle T(n,k), read by rows, given by (1,0,2,1,3,2,4,3,5,4,6,5,7,6,8,7,9,8,...) DELTA (0,1,0,1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 13, 9, 1, 0, 1, 46, 56, 16, 1, 0, 1, 199, 334, 160, 25, 1, 0, 1, 1072, 2157, 1408, 365, 36, 1, 0, 1, 6985, 15701, 12445, 4417, 721, 49, 1, 0, 1, 53218, 129214, 116698, 50944, 11452, 1288, 64, 1, 0, 1, 462331, 1191336, 1183216, 597026, 166716, 25956, 2136, 81, 1, 0
Offset: 0
Triangle begins :
1
1, 0
1, 1, 0
1, 4, 1, 0
1, 13, 9, 1, 0
1, 46, 56, 16, 1, 0
1, 199, 334, 160, 25, 1, 0
1, 1072, 2157, 1408, 365, 36, 1, 0
1, 6985, 15701, 12445, 4417, 721, 49, 1, 0
1, 53218, 129214, 116698, 50944, 11452, 1288, 64, 1, 0
-
DELTA[r_, s_, m_] := Module[{p, q, t, x, y}, q[k_] := x*r[[k + 1]] + y*s[[k + 1]]; p[0, ] = 1; p[, -1] = 0; p[n_ /; n >= 1, k_ /; k >= 0] := p[n, k] = p[n, k - 1] + q[k]*p[n - 1, k + 1] // Expand; t[n_, k_] := Coefficient[p[n, 0], x^(n - k)*y^k]; t[0, 0] = p[0, 0]; Table[t[n, k], {n, 0, m}, {k, 0, n}]];
m = 10;
DELTA[LinearRecurrence[{1, 1, -1}, {1, 0, 2}, m], LinearRecurrence[{0, 1}, {0, 1}, m], m] // Flatten (* Jean-François Alcover, Feb 21 2019 *)
A288528
Numbers with consecutive positive decimal digits after the digits are sorted.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 123, 132, 213, 231, 234, 243, 312, 321, 324, 342, 345, 354, 423, 432, 435, 453, 456, 465, 534, 543, 546, 564, 567, 576, 645, 654, 657, 675, 678, 687, 756, 765, 768, 786, 789, 798, 867, 876, 879, 897, 978, 987
Offset: 1
-
def ok(n): return "".join(sorted(str(n))) in "123456789"
print([k for k in range(999) if ok(k)]) # Michael S. Branicky, Aug 04 2022
-
# alternate for generating full sequence instantly
from itertools import permutations
frags = ["123456789"[i:j] for i in range(9) for j in range(i+1, 10)]
afull = sorted(int("".join(s)) for f in frags for s in permutations(f))
print(afull[:70]) # Michael S. Branicky, Aug 04 2022
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