A142150 -id:A142150 - cp's OEIS frontend

A237719 Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).

1, 2, 6, 12, 18, 20, 24, 28, 30, 40, 42, 54, 56, 66, 70, 78, 80, 88, 100, 102, 104, 112, 114, 120, 126, 138, 140, 150, 160, 162, 174, 176, 180, 186, 196, 198, 200, 204, 208, 220, 222, 224, 228, 234, 240, 246, 258, 260, 272, 276, 282, 294, 304, 306, 308, 318, 320

Offset: 1

Jaroslav Krizek, Mar 16 2014

nonn

Numbers n such that k(n) = A142150(n) = A229110(n) + A054024(n).
Numbers n such that k(n) = (A000217(n) mod n) = (A024816(n) mod n) + (A000203(n) mod n).
k(n) = 0 for odd n, k(n) = n/2 for even n.
If there are any odd multiply-perfect numbers, they are members of this sequence.
If there is no odd multiply-perfect number, then:
(1) the only odd number in this sequence is 1,
(2) corresponding sequence of numbers k(n): {0; a(n) / 2 for n > 1}.
Supersequence of A159907, A007691 and A000396.

			12 is in the sequence because k(12) = (12*(12+1)/2) mod 12 = antisigma(12) mod 12 + sigma(12) mod 12; k(12) = 6 = 4 + 2 = n/2.

Cf. A000203, A000217, A024816, A054024, A142150, A229110.

[n: n in [1..320] | IsZero(n*(n+1)div 2 mod n - SumOfDivisors(n) mod n - (n*(n+1)div 2-SumOfDivisors(n)) mod n)]

A257846 a(n) = floor(n/6) * (n mod 6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 0, 2, 4, 6, 8, 10, 0, 3, 6, 9, 12, 15, 0, 4, 8, 12, 16, 20, 0, 5, 10, 15, 20, 25, 0, 6, 12, 18, 24, 30, 0, 7, 14, 21, 28, 35, 0, 8, 16, 24, 32, 40, 0, 9, 18, 27, 36, 45, 0, 10, 20, 30, 40, 50, 0, 11, 22, 33, 44, 55, 0, 12, 24

Offset: 0

M. F. Hasler, May 10 2015

Equivalently, write n in base 6, multiply the last digit by the number with its last digit removed.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A142150 (the base 2 analog), A115273, A257844 - A257850.

Table[Floor[n/6]*Mod[n, 6], {n, 120}] (* Michael De Vlieger, May 11 2015 *)

PARI
```
a(n,b=6)=(n=divrem(n,b))[1]*n[2]
```

concat([0, 0, 0, 0, 0, 0, 0], Vec(x^7*(5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2) + O(x^100))) \\ Colin Barker, May 11 2015

a(n) = 2*a(n-6)-a(n-12). - Colin Barker, May 11 2015

G.f.: x^7*(5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2). - Colin Barker, May 11 2015

A257847 a(n) = floor(n/7) * (n mod 7).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 0, 2, 4, 6, 8, 10, 12, 0, 3, 6, 9, 12, 15, 18, 0, 4, 8, 12, 16, 20, 24, 0, 5, 10, 15, 20, 25, 30, 0, 6, 12, 18, 24, 30, 36, 0, 7, 14, 21, 28, 35, 42, 0, 8, 16, 24, 32, 40, 48, 0, 9, 18, 27, 36, 45, 54, 0, 10, 20

Offset: 0

M. F. Hasler, May 10 2015

Equivalently, write n in base 7, multiply the last digit by the number with its last digit removed.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,2,0,0,0,0,0,0,-1).

Cf. A194757.

Cf. A142150 (the base 2 analog), A115273, A257844 - A257850.

Table[Floor[n/7]Mod[n,7],{n,0,80}] (* Harvey P. Dale, Nov 12 2022 *)

PARI
```
a(n,b=7)=(n=divrem(n,b))[1]*n[2]
```

concat([0,0,0,0,0,0,0,0], Vec(x^8*(6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x^6+x^5+x^4+x^3+x^2+x+1)^2) + O(x^100))) \\ Colin Barker, May 11 2015

a(n) = 2*a(n-7)-a(n-14). - Colin Barker, May 11 2015

G.f.: x^8*(6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x^6+x^5+x^4+x^3+x^2+x+1)^2). - Colin Barker, May 11 2015

A257848 a(n) = floor(n/8) * (n mod 8).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 0, 2, 4, 6, 8, 10, 12, 14, 0, 3, 6, 9, 12, 15, 18, 21, 0, 4, 8, 12, 16, 20, 24, 28, 0, 5, 10, 15, 20, 25, 30, 35, 0, 6, 12, 18, 24, 30, 36, 42, 0, 7, 14, 21, 28, 35, 42, 49, 0, 8, 16, 24, 32, 40, 48, 56, 0, 9

Offset: 0

M. F. Hasler, May 10 2015

Equivalently, write n in base 8, multiply the last digit by the number with its last digit removed.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1).

Cf. A142150 (the base 2 analog), A115273, A257844 - A257850.

Table[Floor[n/8]Mod[n,8],{n,0,90}] (* or *) LinearRecurrence[{0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1},{0,0,0,0,0,0,0,0,0,1,2,3,4,5,6,7},90] (* Harvey P. Dale, Nov 05 2023 *)

PARI
```
a(n,b=8)=(n=divrem(n,b))[1]*n[2]
```

concat([0,0,0,0,0,0,0,0,0], Vec(x^9*(7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)^2) + O(x^100))) \\ Colin Barker, May 11 2015

def A257848(n): return (n>>3)*(n&7) # Chai Wah Wu, Jan 19 2023

a(n) = 2*a(n-8)-a(n-16). - Colin Barker, May 11 2015

G.f.: x^9*(7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)^2). - Colin Barker, May 11 2015

A359329 Number of diagonals in a regular polygon with n sides not passing through the center.

Original entry on oeis.org

0, 0, 5, 6, 14, 16, 27, 30, 44, 48, 65, 70, 90, 96, 119, 126, 152, 160, 189, 198, 230, 240, 275, 286, 324, 336, 377, 390, 434, 448, 495, 510, 560, 576, 629, 646, 702, 720, 779, 798, 860, 880, 945, 966, 1034, 1056, 1127, 1150, 1224, 1248, 1325, 1350, 1430, 1456, 1539, 1566, 1652, 1680

Offset: 3

Luk De Clercq, Dec 26 2022

Paolo Xausa, Table of n, a(n) for n = 3..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

A014106 and A054000 interleaved.

Cf. A000096, A142150.

Table[(n*(n - 4 + BitGet[n, 0]))/2, {n, 3, 100}] (* Paolo Xausa, Oct 02 2024 *)

def A359329(n): return (n*(n-4)+n*(n&1))>>1 # Chai Wah Wu, Jan 23 2023

If n is odd, a(n) = (n^2 - 3*n)/2; if n is even, a(n) = (n^2 - 4*n)/2.
a(n) = A000096(n-3) - A142150(n-3).
G.f.: x^5*(5 + x - 2*x^2)/((1 - x)^3*(1 + x)^2). - Stefano Spezia, Jan 04 2023

A367205 Number of nonnegative sequences of integers S with the properties that (1) sum(S) + length(S) = n and (2) there exists a nonnegative sequence whose Euler transform begins with S starting at index 1.

Original entry on oeis.org

1, 2, 3, 6, 9, 16, 27, 45, 74, 125, 205, 343, 564, 934, 1535, 2536, 4165, 6855, 11249, 18465

Offset: 1

Peter Kagey, Nov 10 2023

a(n) <= 2^(n-1), which is the number of nonnegative sequences S with sum(S) + length(S) = n.

The candidate sequences are related to the row n of A228369, by subtracting 1 from each term.

			For n = 4 the a(4) = 6 sequences are
1) (0,0,0,0) because (1,0,0,0,0,...) = Euler(0,0,0,0,...),
2) (0,0,1)   because (1,0,0,1,...)   = Euler(0,0,1,...),
3) (0,1,0)   because (1,0,1,0,...)   = Euler(0,1,0,...),
4) (0,2)     because (1,0,2,...)     = Euler(0,2,...),
5) (1,1)     because (1,1,1,...)     = Euler(1,0,...) and
6) (3)       because (1,3,...)       = Euler(3,...).
The lexicographically earliest such sequences are:
1) A000007 = Euler(0,0,0,0,...)
2) A079978 = Euler(0,0,1,0,...)
3) A000035 = Euler(0,1,0,0,...)
4) A142150 = Euler(0,2,0,0,...)
5) A000012 = Euler(1,0,0,0,...)
6) A000217 = Euler(3,0,0,0,...)
Note that (2,0) and (1,0,0) are not the 1-indexed prefix of the Euler transform of a nonnegative sequence.

Cf. A000007, A000012, A000035, A000217, A079978, A142150

Cf. A228369.

A367205[n_] :=
 Select[EulerInvTransform /@ (Map[# - 1 &, #] & /@
     Join @@ Permutations /@ IntegerPartitions[n]),
  AllTrue[#, # >= 0 &] &]

A377825 Number of distinct permutations of the terms of the n-th row of Pascal's triangle with alternating signs.

Original entry on oeis.org

1, 2, 3, 24, 30, 720, 630, 40320, 22680, 3628800, 1247400, 479001600, 97297200, 87178291200, 10216206000, 20922789888000, 1389404016000, 6402373705728000, 237588086736000, 2432902008176640000, 49893498214560000, 1124000727777607680000, 12623055048283680000

Offset: 0

Ryan Jean, Nov 08 2024

Note that for any given n, there are n+1 terms in that row.

			For n = 0, a(0) = 1 since there is just one term.
For n = 1, the signed row terms are {1, -1} so a(1) = 2 permutations.
For n = 2, the signed row terms are {1, -2, 1} which have only a(2) = 3 distinct permutations.
For n = 3, the signed row terms are {1, -3, 3, -1} which have a(3) = 24 permutations.

Paolo Xausa, Table of n, a(n) for n = 0..400

Bisections are: A007019, A010050.

Cf. A007318, A072345, A142150.

seq((n+1)! / (2^((n*(1+(-1)^n)) / 4)), n=0..22); # Georg Fischer, Dec 19 2024

A377825[n_] := (n+1)!/2^((n*(1 + (-1)^n))/4); Array[A377825, 25, 0] (* Paolo Xausa, Dec 20 2024 *)

a(n) = (n+1)! / (2^((n*(1+(-1)^n)) / 4)).
E.g.f.: 2*(x^6+x^5-4*x^3-3*x^2+4*x+2)/((x-1)^2*(x+1)^2*(x^2-2)^2). - Alois P. Heinz, Nov 09 2024
a(n) = (n+1)!/A072345(n-1) for n > 0. - Stefano Spezia, Nov 09 2024
Sum_{n>=0} 1/a(n) = cosh(1) + sinh(sqrt(2))/sqrt(2) - 1. - Amiram Eldar, Dec 25 2024

a(22) corrected by Georg Fischer, Dec 19 2024

A316744 a(n) is the smallest number having exactly n ways to be represented as sum of at least two consecutive positive integers and expressible as sum of n consecutive positive integers, or 0 if no such number exists.

Original entry on oeis.org

9, 15, 162, 45, 729, 105, 900, 405, 9765625, 495, 1062882, 9477, 3969, 945, 344373768, 3825, 387420489, 7695, 34650, 413343, 81402749386839761113321, 7245, 202500, 732421875, 38025, 25515, 919973073089479921953602, 58725, 0, 29295, 23619600, 473513931, 60886809, 17325, 300189270593998242

Offset: 2

Jianing Song, Jul 13 2018

nonn

a(n) is the smallest number such that A069283(n) == n*(n-1)/2 == A142150(n) (mod n). Or equivalently, a(n) is the smallest number of the form 2^t*s, where s is an odd number with exactly n + 1 divisors and divisible by A000265(n), t = 0 for odd n and A007814(n) - 1 for even n.
Let n = 2^e_0*Product_{i=1..m} p_i^e_i where p_i are odd primes; n + 1 = Product_{j=1..s} q_j where q_j are primes, then a(n) != 0 iff there is an injection f from {1,2,..,m} to {1,2,...,s} such that q_f(i) >= e_i + 1 for all 1 <= i <= m, implying s >= m. If such an injection does exist, then the number of k having exactly n ways to be represented as sum of at least two consecutive positive integers and expressible as sum of n consecutive positive integers is finite iff s = m, in which case the number of k is equal to the number of injections such that if i < j and e_i = e_j then q_f(i) <= q_f(j).
If A038547(n) is divisible by A000265(n), then a(n) = 2^t*A038547(n), t defined as above.
If n + 1 is a Fermat prime, then a(n) = (n/2)*3^n. If n = p - 1 = 2^e*q with p, q primes, then a(n) = 2^(e-1)*q^n, which is relatively large.

			a(2) = 9 = 4 + 5 = 2 + 3 + 4.
a(3) = 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.
a(4) = 162 = 53 + 54 + 55 = 39 + 40 + 41 + 42 = 14 + 15 + 16 + ... + 22 = 8 + 9 + 10 + ... + 19.
If a number k has exactly 30 ways to be represented as sum of at least two consecutive positive integers, then it must have exactly 31 odd divisors. On the other hand, the sum of 30 consecutive positive integers is congruent to 15 mod 30, so k must be of the form p^30 where p is an odd prime, which obviously cannot be divisible by 15. So a(30) = 0.
Let n = 225 = 3^2*5^2, n + 1 = 226 = 2*113, so e_1 = 2, e_2 = 2, q_1 = 2, q_2 = 113. An injection from {1,2} to {1,2} such that q_f(1) >= e_1 + 1 and q_f(2) >= e_2 + 1 does not exist, so a(225) = 0.

Cf. A000265, A001222, A005087, A007814, A038547, A069283, A142150.

cp's OEIS Frontend

A237719 Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).

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A257846 a(n) = floor(n/6) * (n mod 6).

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A257847 a(n) = floor(n/7) * (n mod 7).

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A257848 a(n) = floor(n/8) * (n mod 8).

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A359329 Number of diagonals in a regular polygon with n sides not passing through the center.

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A367205 Number of nonnegative sequences of integers S with the properties that (1) sum(S) + length(S) = n and (2) there exists a nonnegative sequence whose Euler transform begins with S starting at index 1.

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A377825 Number of distinct permutations of the terms of the n-th row of Pascal's triangle with alternating signs.

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A316744 a(n) is the smallest number having exactly n ways to be represented as sum of at least two consecutive positive integers and expressible as sum of n consecutive positive integers, or 0 if no such number exists.