A102287 -id:A102287 - cp's OEIS frontend

A069514 Numbers n such that sigma(reversal(n)) = reversal(sigma(n)). Ignore leading 0's.

1, 2, 3, 4, 5, 7, 14, 41, 124, 194, 333, 421, 491, 1324, 4231, 13324, 17054, 17571, 42331, 45071, 120530, 138465, 386650, 564831, 1130324, 1216360, 1333324, 1727571, 1757271, 1757571, 1787871, 2249422, 4230311, 4233331, 4369634

Offset: 1

Joseph L. Pe, Apr 15 2002

For an arithmetical function f, call the arguments n such that f(reverse(n)) = reverse(f(n)) the "palinpoints" of f. This sequence is the sequence of palinpoints of f(n) = sigma(n).
If n is in the sequence and 10 doesn't divide n then the reversal of n is also in the sequence. - Farideh Firoozbakht, Aug 31 2004
Comments from Farideh Firoozbakht, Jan 16 2005. "The largest term that I found is M=(58*100^687 - 157)/33; the length of M is 1375. I proved the following facts about this sequence:
"I : If p=(58*100^n - 157)/99 is prime then 3*p is in the sequence, the sequence A102285 gives such n's.
"II : If p=(59*100^n - 257)/99 is prime then 3*p is in the sequence, I found only two primes of this form the first for n=3 and the second for n=27, next such n is greater than 3400.
"III : If both numbers p=10^n - 3 & q=5*10^n - 9 are primes then both numbers 2*p & q are in the sequence, q is reversal of 2*p. I found only two such n's, n=1 & 2.
"IV : If both numbers p=(10^n-7)/3 & q=(127*10^(n-1)-7)/3 are primes then both numbers 4*p & q are in the sequence, q is the reversal of 4*p, the sequence A102287 are these terms of A069514, I found only four such n's, n=2,3,4 & 6."

			Let f(n) = sigma(n). Then f(194) = 294, f(491) = 492, so f(reverse(194)) = reverse(f(194)). Therefore 194 belongs to the sequence.

Cf. A102285, A102286, A102287.

rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := DivisorSigma[1, n]; Select[Range[10^6], f[rev[ # ]] == rev[f[ # ]] &]

More terms from Farideh Firoozbakht, Aug 31 2004

A124322 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} (or of any n-set) having k blocks of even size (0<=k<=floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 7, 3, 12, 25, 15, 37, 91, 60, 15, 128, 329, 315, 105, 457, 1415, 1533, 630, 105, 1872, 6297, 7623, 4410, 945, 8169, 29431, 42150, 27405, 7875, 945, 37600, 151085, 233475, 176715, 69300, 10395, 188685, 802099, 1365243, 1199220, 533610

Offset: 0

Emeric Deutsch, Oct 28 2006

Row n has 1+floor(n/2) terms. Sum of row n is the Bell number B(n)=A000110(n). Sum_{k=0..floor(n/2)} k*T(n,k) = A102287(n). T(n,0)=A003724(n).

			T(4,1) = 7 because we have 1234, 14|2|3, 1|24|3, 1|2|34, 13|2|4, 1|23|4 and 12|3|4.
Triangle starts:
   1;
   1;
   1,  1;
   2,  3;
   5,  7,  3;
  12, 25, 15;
  37, 91, 60, 15;
  ...

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 225.

Alois P. Heinz, Rows n = 0..200, flattened

Cf. A000110, A102287, A003724, A124321.

G:=exp(sinh(z)+t*(cosh(z)-1)): Gser:=simplify(series(G,z=0,16)): for n from 0 to 13 do P[n]:=sort(n!*coeff(Gser,z,n)) od: for n from 0 to 13 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form
# second Maple program:
with(combinat):
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
      `if`(i<1, 0, add(multinomial(n, n-i*j, i$j)/j!*
      b(n-i*j, i-1)*`if`(irem(i, 2)=0, x^j, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..15);  # Alois P. Heinz, Mar 08 2015

nn = 10; Range[0, nn]! CoefficientList[Series[Exp[y (Cosh[x] - 1) + Sinh[x]], {x, 0, nn}], {x, y}] // Grid  (* Geoffrey Critzer, Aug 28 2012*)

E.g.f.: exp[sinh(z)+t(cosh(z)-1)].

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A069514 Numbers n such that sigma(reversal(n)) = reversal(sigma(n)). Ignore leading 0's.

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A124322 Triangle read by rows: T(n,k) is the number of set partitions of {1,2,...,n} (or of any n-set) having k blocks of even size (0<=k<=floor(n/2)).

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