A236857 -id:A236857 - cp's OEIS frontend

A236856 Partial sums of A003418 starting summing from A003418(1), with a(0) = 0.

0, 1, 3, 9, 21, 81, 141, 561, 1401, 3921, 6441, 34161, 61881, 422241, 782601, 1142961, 1863681, 14115921, 26368161, 259160721, 491953281, 724745841, 957538401, 6311767281, 11665996161, 38437140561, 65208284961, 145521718161, 225835151361, 2554924714161

Offset: 0

Antti Karttunen, Feb 27 2014

nonn

Similar comments about the trailing digits apply here as in A173185.
a(n) gives the position of the last element of row n in irregular tables like A238280.
From a(2)=3 onward all terms are divisible by three.
a(n) is divisible by 73 for n >= 72. Therefore a(n)/3 is prime for only 13 values of n: 3, 4, 6, 8, 9, 12, 16, 22, 23, 31, 35, 48 and 53. - Amiram Eldar, Sep 19 2022

Antti Karttunen, Table of n, a(n) for n = 0..250

One less than A173185.

Cf. A003418, A007489, A231722, A236857, A236858.

Prepend[Accumulate @ Table[LCM @@ Range[n], {n, 1, 30}], 0] (* Amiram Eldar, Sep 19 2022 *)

(define (A236856 n) (if (< n 2) n (+ (A236856 (- n 1)) (A003418 n))))

a(n) = A173185(n)-1.

A236858 Irregular table where row n contains numbers from 1 to the least common multiple (LCM) of {1, 2, ..., n}. Row 0 is given as a(0)=1.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60

Offset: 0

Antti Karttunen, Feb 27 2014

Numbers 1..A003418(n) followed by numbers 1..A003418(n+1), etc., where A003418(n) gives the least common multiple (LCM) of {1, 2, ..., n} for n >= 1 and A003418(0)=1.

Useful when computing irregular tables like A238280. Note that as A238280 begins with row 1, it starts referring to this sequence only from a(1) onward.

			The sequence can be viewed also as an irregular table that starts as:
0 | 1;
1 | 1;
2 | 1, 2;
3 | 1, 2, 3, 4, 5, 6;
4 | 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12;
...

Antti Karttunen, Rows 0..10 of the table, flattened

A003418(n) gives the last term of each row n.

Cf. A220661, A236856, A236857, A238280.

Join[{1},Table[Range[LCM@@Range[n]],{n,6}]//Flatten] (* Harvey P. Dale, Apr 29 2019 *)

(define (A236858 n) (- n (A236856 (- (A236857 n) 1))))

a(n) = n - A236856(A236857(n)- 1).

A238280 Irregular triangle read by rows, T(n,k) = Sum_{i = 1..n} k mod i, k = 1..m where m = lcm(1..n).

Original entry on oeis.org

0, 1, 0, 2, 2, 1, 1, 3, 0, 3, 4, 4, 1, 4, 2, 5, 2, 2, 3, 6, 0, 4, 6, 7, 5, 4, 3, 7, 5, 6, 3, 7, 2, 6, 8, 4, 2, 6, 5, 9, 2, 3, 5, 9, 4, 3, 5, 6, 4, 8, 2, 6, 4, 5, 7, 6, 1, 5, 7, 8, 1, 5, 4, 8, 6, 2, 4, 8, 3, 7, 4, 5, 3, 7, 6, 5, 3, 4, 6, 10, 0, 5, 8, 10, 9, 9, 3, 8, 7, 9, 7, 12, 2, 7, 10, 7, 6, 11, 5, 10, 4, 6, 9, 14, 4, 4, 7, 9, 8, 13, 2, 7, 6, 8, 11

Offset: 1

Kival Ngaokrajang, Feb 22 2014

Row n contains A003418(n) terms.

The penultimate term (the one before zero) of row n = A000217(n-1).

			Row n of this irregular triangle is obtained by taking the first A003418(n) = lcm(1..n) terms (up to and including the first zero) of the following array (which starts at row n=1 and column k=1 and is periodic in each row):
  0; 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
  1  0; 1  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0
  2  2  1  1  3  0; 2  2  1  1  3  0  2  2  1  1  3  0  2  2 # A110269
  3  4  4  1  4  2  5  2  2  3  6  0; 3  4  4  1  4  2  5  2
  4  6  7  5  4  3  7  5  6  3  7  2  6  8  4  2  6  5  9  2
  5  8 10  9  9  3  8  7  9  7 12  2  7 10  7  6 11  5 10  4
  6 10 13 13 14  9  8  8 11 10 16  7 13 10  8  8 14  9 15 10
  7 12 16 17 19 15 15  8 12 12 19 11 18 16 15  8 15 11 18 14
  8 14 19 21 24 21 22 16 12 13 21 14 22 21 21 15 23 11 19 16
  9 16 22 25 29 27 29 24 21 13 22 16 25 25 26 21 30 19 28 16

Antti Karttunen, Rows 1..10 of the table, flattened
Kival Ngaokrajang, Illustration for n = 1..10

Cf. A000217, A003418, A173185, A236856, A236857, A236858.

(define (A238280 n) (A238280tabf (A236857 n) (A236858 n)))
(define (A238280tabf n k) (add (lambda (i) (modulo k i)) 1 n))
;; Implements sum_{i=lowlim..uplim} intfun(i):
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))
;; Antti Karttunen, Feb 27 2014

A173185 Partial sums of A003418.

Original entry on oeis.org

1, 2, 4, 10, 22, 82, 142, 562, 1402, 3922, 6442, 34162, 61882, 422242, 782602, 1142962, 1863682, 14115922, 26368162, 259160722, 491953282, 724745842, 957538402, 6311767282, 11665996162, 38437140562, 65208284962, 145521718162, 225835151362, 2554924714162

Offset: 0

Jonathan Vos Post, Feb 12 2010

From Antti Karttunen, Feb 27 2014: (Start)
For all n >= 4, a(n) mod 10 = 2 (as A003418(5) = 60, the first multiple of ten in that sequence).
For all n >= 24, a(n) mod 100 = 62 (as A003418(25) = 26771144400, the first multiple of one hundred in that sequence).
Cf. also A236856.
a(n-1) gives the position of the first element of row n in irregular tables like A238280.
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..2297 (first 251 terms from Antti Karttunen)

One more than A236856.

Cf. A002944, A003418, A003422, A102910, A093880, A133233, A231721, A236857, A236858, A238280.

b:= proc(n) b(n):= `if`(n=0, 1, ilcm(n, b(n-1))) end:
a:= proc(n) a(n):= `if`(n<0, 0, a(n-1) +b(n)) end:
seq(a(n), n=0..35);  # Alois P. Heinz, Mar 31 2018

Table[If[n == 0, 1, LCM @@ Range[n]], {n, 0, 50}] // Accumulate (* Jean-François Alcover, Jan 03 2022 *)

a(n) = sum(k=0, n, lcm(vector(k, i, i))); \\ Michel Marcus, Mar 13 2018

(define (A173185 n) (if (< n 1) 1 (+ (A173185 (- n 1)) (A003418 n))))

a(n) = Sum_{i=0..n} A003418(i).

Missing term a(9)=3922 inserted by Antti Karttunen, Feb 27 2014

cp's OEIS Frontend

A236856 Partial sums of A003418 starting summing from A003418(1), with a(0) = 0.

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A236858 Irregular table where row n contains numbers from 1 to the least common multiple (LCM) of {1, 2, ..., n}. Row 0 is given as a(0)=1.

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A238280 Irregular triangle read by rows, T(n,k) = Sum_{i = 1..n} k mod i, k = 1..m where m = lcm(1..n).

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A173185 Partial sums of A003418.

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