A001923 -id:A001923 - cp's OEIS frontend

A348014 Triangle, read by rows, with row n forming the coefficients in Product_{k=0..n} (1 + k^k*x).

1, 1, 1, 1, 5, 4, 1, 32, 139, 108, 1, 288, 8331, 35692, 27648, 1, 3413, 908331, 26070067, 111565148, 86400000, 1, 50069, 160145259, 42405161203, 1216436611100, 5205269945088, 4031078400000

Offset: 0

Seiichi Manyama, Sep 24 2021

			Triangle begins:
  1;
  1,    1;
  1,    5,      4;
  1,   32,    139,      108;
  1,  288,   8331,    35692,     27648;
  1, 3413, 908331, 26070067, 111565148, 86400000;

Seiichi Manyama, Rows n = 0..37, flattened

Column k=1 gives A001923.
The diagonal of the triangle is A002109.
Cf. A007318, A008955, A023531, A108084, A173007, A173008, A249677.

T(n, k) = if(k==0, 1, if(k==n, prod(j=1, n, j^j), T(n-1, k)+n^n*T(n-1, k-1)));

PARI
```
row(n) = Vecrev(prod(k=1, n, 1+k^k*x));
```

T(0,0) = 1; T(n,k) = T(n-1,k) + n^n * T(n-1,k-1).

A383228 a(n) is the number of cases where both j and k (1 <= j < k <= n), are divisors of Sum_{i=j..k} i^i.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17

Offset: 1

Jean-Marc Rebert, Apr 20 2025

nonn

			1 and 4 divide Sum_{i = 1..4} i^i = 288, and
1 and 17 divide Sum_{i = 1..17} i^i = 846136323944176515621, and
1 and 19 divide Sum_{i = 1..19} i^i = 2018612200059554303215024, and
2 and 9 divide Sum_{i = 2..9} i^i = 405071316, and
2 and 30 divide Sum_{i = 2..30} i^i = 208492413443704093346554910065262730566475780, and
3 and 6 divide Sum_{i = 3..6} i^i = 50064, and
5 and 15 divide Sum_{i = 5..15} i^i = 449317984130199540, and
14 and 42 divide Sum_{i = 14..42} i^i = 151474018115847331407847533862930150275321445330384244581082952733296, and
22 and 26 divide Sum_{i = 22..26} i^i = 6246292379849897330286605654999947328, so a(42) = 9.

Jean-Marc Rebert, Table of n, a(n) for n = 1..1000

Cf. A000312, A001923, A128981.

a(n) = sum(j=1, n, sum(k=j+1, n, my(s=sum(i=j, k, i^i)); !(s % j) && !(s % k))); \\ Michel Marcus, Apr 20 2025

More terms from Michel Marcus, Apr 20 2025

A319515 Second term of the simple continued fraction of (Sum_{k=1..n} k^k)/(n^n).

Original entry on oeis.org

1, 4, 5, 8, 10, 13, 16, 19, 21, 24, 27, 30, 32, 35, 38, 41, 43, 46, 49, 51, 54, 57, 60, 62, 65, 68, 70, 73, 76, 79, 81, 84, 87, 90, 92, 95, 98, 100, 103, 106, 109, 111, 114, 117, 119, 122, 125, 128, 130, 133, 136, 138, 141, 144, 147, 149, 152, 155, 158, 160, 163, 166, 168, 171, 174, 177

Offset: 1

Greg Huber, Sep 21 2018

Ignoring the first two terms, a(n+1) is very similar to floor(n*e) (the Beatty sequence for e, A022843). This breaks down at n=11, where a(12)=30 and floor(11*e)=29. The successive terms (or partial denominators) in the sequence differ by 3 except for semi-regular "glitches". For example, the pattern of differences is 313233323332333233233323323332333233233323323332332333233323323332... A natural conjecture is that the mean of these differences is asymptotic to e. The glitches are positioned between terms n and n+1, for values of n which begin 2,4,8,12,16,19,23,26,30,34,37,41,44,48,.... If one counts the 3's immediately preceding the 2's in the pattern of differences, one gets the "derivative" sequence 233323233232323323....

			a(1)=1 because the continued fraction of 1 is written as 0 + 1/1 = (0;1).
a(3)=5 because (1 + 4 + 27)/27 = 1 + 5/27 = 1 + 1/(5 + 1/(2 + 1/2)) = (1;5,2,2).

G. W. Wishard and F. Underwood, Problem 4155, Amer. Math. Monthly, 53 (1946), 471.

Cf. A001923, A022843.

a[1]=1; a[n_] := ContinuedFraction[Sum[k^k, {k, 1, n}]/n^n][[2]]; Array[a, 100]

a(n) = if (n==1, 1, contfrac(sum(k=1, n, k^k)/n^n)[2]); \\ Michel Marcus, Sep 23 2018

A348361 a(n) = Sum_{k=1..n} k^(k'), where ' is the arithmetic derivative.

Original entry on oeis.org

1, 3, 6, 262, 267, 8043, 8050, 68719484786, 68720016227, 68730016227, 68730016238, 184884327625052654, 184884327625052667, 184884348286099451, 184884350848990076, 340282366920938463463559491782617201532, 340282366920938463463559491782617201549, 340282366921167931715454621189757074317

Offset: 1

Wesley Ivan Hurt, Oct 14 2021

nonn

			a(4) = 262; a(4) = Sum_{k=1..4} k^(k') = 1^(1') + 2^(2') + 3^(3') + 4^(4') = 1^0 + 2^1 + 3^1 + 4^4 = 262.

Cf. A001923, A003415.

a:= n-> add(k^(k*add(i[2]/i[1], i=ifactors(k)[2])), k=1..n):
seq(a(n), n=1..18);  # Alois P. Heinz, Oct 14 2021

Accumulate@ Array[#^If[# < 2, 0, # Total[#2/#1 & @@@ FactorInteger[#]]] &, 18] (* Michael De Vlieger, Oct 14 2021 *)

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A348014 Triangle, read by rows, with row n forming the coefficients in Product_{k=0..n} (1 + k^k*x).

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A383228 a(n) is the number of cases where both j and k (1 <= j < k <= n), are divisors of Sum_{i=j..k} i^i.

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A319515 Second term of the simple continued fraction of (Sum_{k=1..n} k^k)/(n^n).

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A348361 a(n) = Sum_{k=1..n} k^(k'), where ' is the arithmetic derivative.

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